Calculus Examples

$f (x) = - 8 e^{- 8 x} + 5 e^{- 5 x}$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $- 8 e^{- 8 x} + 5 e^{- 5 x}$ with respect to $x$ is $\frac{d}{d x} [- 8 e^{- 8 x}] + \frac{d}{d x} [5 e^{- 5 x}]$ .

$\frac{d}{d x} [- 8 e^{- 8 x}] + \frac{d}{d x} [5 e^{- 5 x}]$

Step 1.2

Evaluate $\frac{d}{d x} [- 8 e^{- 8 x}]$ .

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Step 1.2.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 e^{- 8 x}$ with respect to $x$ is $- 8 \frac{d}{d x} [e^{- 8 x}]$ .

$- 8 \frac{d}{d x} [e^{- 8 x}] + \frac{d}{d x} [5 e^{- 5 x}]$

Step 1.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 8 x$ .

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Step 1.2.2.1

To apply the Chain Rule, set $u_{1}$ as $- 8 x$ .

$- 8 (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [- 8 x]) + \frac{d}{d x} [5 e^{- 5 x}]$

Step 1.2.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$- 8 (e^{u_{1}} \frac{d}{d x} [- 8 x]) + \frac{d}{d x} [5 e^{- 5 x}]$

Step 1.2.2.3

Replace all occurrences of $u_{1}$ with $- 8 x$ .

$- 8 (e^{- 8 x} \frac{d}{d x} [- 8 x]) + \frac{d}{d x} [5 e^{- 5 x}]$

Step 1.2.3

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 x$ with respect to $x$ is $- 8 \frac{d}{d x} [x]$ .

$- 8 (e^{- 8 x} (- 8 \frac{d}{d x} [x])) + \frac{d}{d x} [5 e^{- 5 x}]$

Step 1.2.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$- 8 (e^{- 8 x} (- 8 \cdot 1)) + \frac{d}{d x} [5 e^{- 5 x}]$

Step 1.2.5

Multiply $- 8$ by $1$ .

$- 8 (e^{- 8 x} \cdot - 8) + \frac{d}{d x} [5 e^{- 5 x}]$

Step 1.2.6

Move $- 8$ to the left of $e^{- 8 x}$ .

$- 8 (- 8 \cdot e^{- 8 x}) + \frac{d}{d x} [5 e^{- 5 x}]$

Step 1.2.7

Multiply $- 8$ by $- 8$ .

$64 e^{- 8 x} + \frac{d}{d x} [5 e^{- 5 x}]$

Step 1.3

Evaluate $\frac{d}{d x} [5 e^{- 5 x}]$ .

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Step 1.3.1

Since $5$ is constant with respect to $x$ , the derivative of $5 e^{- 5 x}$ with respect to $x$ is $5 \frac{d}{d x} [e^{- 5 x}]$ .

$64 e^{- 8 x} + 5 \frac{d}{d x} [e^{- 5 x}]$

Step 1.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 5 x$ .

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Step 1.3.2.1

To apply the Chain Rule, set $u_{2}$ as $- 5 x$ .

$64 e^{- 8 x} + 5 (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- 5 x])$

Step 1.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$64 e^{- 8 x} + 5 (e^{u_{2}} \frac{d}{d x} [- 5 x])$

Step 1.3.2.3

Replace all occurrences of $u_{2}$ with $- 5 x$ .

$64 e^{- 8 x} + 5 (e^{- 5 x} \frac{d}{d x} [- 5 x])$

Step 1.3.3

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x$ with respect to $x$ is $- 5 \frac{d}{d x} [x]$ .

$64 e^{- 8 x} + 5 (e^{- 5 x} (- 5 \frac{d}{d x} [x]))$

Step 1.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$64 e^{- 8 x} + 5 (e^{- 5 x} (- 5 \cdot 1))$

Step 1.3.5

Multiply $- 5$ by $1$ .

$64 e^{- 8 x} + 5 (e^{- 5 x} \cdot - 5)$

Step 1.3.6

Move $- 5$ to the left of $e^{- 5 x}$ .

$64 e^{- 8 x} + 5 (- 5 \cdot e^{- 5 x})$

Step 1.3.7

Multiply $- 5$ by $5$ .

$64 e^{- 8 x} - 25 e^{- 5 x}$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $64 e^{- 8 x} - 25 e^{- 5 x}$ with respect to $x$ is $\frac{d}{d x} [64 e^{- 8 x}] + \frac{d}{d x} [- 25 e^{- 5 x}]$ .

$f'' (x) = \frac{d}{d x} (64 e^{- 8 x}) + \frac{d}{d x} (- 25 e^{- 5 x})$

Step 2.2

Evaluate $\frac{d}{d x} [64 e^{- 8 x}]$ .

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Step 2.2.1

Since $64$ is constant with respect to $x$ , the derivative of $64 e^{- 8 x}$ with respect to $x$ is $64 \frac{d}{d x} [e^{- 8 x}]$ .

$f'' (x) = 64 \frac{d}{d x} (e^{- 8 x}) + \frac{d}{d x} (- 25 e^{- 5 x})$

Step 2.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 8 x$ .

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Step 2.2.2.1

To apply the Chain Rule, set $u_{1}$ as $- 8 x$ .

$f'' (x) = 64 (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (- 8 x)) + \frac{d}{d x} (- 25 e^{- 5 x})$

Step 2.2.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$f'' (x) = 64 (e^{u_{1}} \frac{d}{d x} (- 8 x)) + \frac{d}{d x} (- 25 e^{- 5 x})$

Step 2.2.2.3

Replace all occurrences of $u_{1}$ with $- 8 x$ .

$f'' (x) = 64 (e^{- 8 x} \frac{d}{d x} (- 8 x)) + \frac{d}{d x} (- 25 e^{- 5 x})$

Step 2.2.3

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 x$ with respect to $x$ is $- 8 \frac{d}{d x} [x]$ .

$f'' (x) = 64 (e^{- 8 x} (- 8 \frac{d}{d x} x)) + \frac{d}{d x} (- 25 e^{- 5 x})$

Step 2.2.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = 64 (e^{- 8 x} (- 8 \cdot 1)) + \frac{d}{d x} (- 25 e^{- 5 x})$

Step 2.2.5

Multiply $- 8$ by $1$ .

$f'' (x) = 64 (e^{- 8 x} \cdot - 8) + \frac{d}{d x} (- 25 e^{- 5 x})$

Step 2.2.6

Move $- 8$ to the left of $e^{- 8 x}$ .

$f'' (x) = 64 (- 8 \cdot e^{- 8 x}) + \frac{d}{d x} (- 25 e^{- 5 x})$

Step 2.2.7

Multiply $- 8$ by $64$ .

$f'' (x) = - 512 e^{- 8 x} + \frac{d}{d x} (- 25 e^{- 5 x})$

Step 2.3

Evaluate $\frac{d}{d x} [- 25 e^{- 5 x}]$ .

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Step 2.3.1

Since $- 25$ is constant with respect to $x$ , the derivative of $- 25 e^{- 5 x}$ with respect to $x$ is $- 25 \frac{d}{d x} [e^{- 5 x}]$ .

$f'' (x) = - 512 e^{- 8 x} - 25 \frac{d}{d x} e^{- 5 x}$

Step 2.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 5 x$ .

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Step 2.3.2.1

To apply the Chain Rule, set $u_{2}$ as $- 5 x$ .

$f'' (x) = - 512 e^{- 8 x} - 25 (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- 5 x))$

Step 2.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$f'' (x) = - 512 e^{- 8 x} - 25 (e^{u_{2}} \frac{d}{d x} (- 5 x))$

Step 2.3.2.3

Replace all occurrences of $u_{2}$ with $- 5 x$ .

$f'' (x) = - 512 e^{- 8 x} - 25 (e^{- 5 x} \frac{d}{d x} (- 5 x))$

Step 2.3.3

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x$ with respect to $x$ is $- 5 \frac{d}{d x} [x]$ .

$f'' (x) = - 512 e^{- 8 x} - 25 (e^{- 5 x} (- 5 \frac{d}{d x} x))$

Step 2.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = - 512 e^{- 8 x} - 25 (e^{- 5 x} (- 5 \cdot 1))$

Step 2.3.5

Multiply $- 5$ by $1$ .

$f'' (x) = - 512 e^{- 8 x} - 25 (e^{- 5 x} \cdot - 5)$

Step 2.3.6

Move $- 5$ to the left of $e^{- 5 x}$ .

$f'' (x) = - 512 e^{- 8 x} - 25 (- 5 \cdot e^{- 5 x})$

Step 2.3.7

Multiply $- 5$ by $- 25$ .

$f'' (x) = - 512 e^{- 8 x} + 125 e^{- 5 x}$

Step 2.4

Reorder terms.

$f'' (x) = 125 e^{- 5 x} - 512 e^{- 8 x}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$64 e^{- 8 x} - 25 e^{- 5 x} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

By the Sum Rule, the derivative of $- 8 e^{- 8 x} + 5 e^{- 5 x}$ with respect to $x$ is $\frac{d}{d x} [- 8 e^{- 8 x}] + \frac{d}{d x} [5 e^{- 5 x}]$ .

$\frac{d}{d x} [- 8 e^{- 8 x}] + \frac{d}{d x} [5 e^{- 5 x}]$

Step 4.1.2

Evaluate $\frac{d}{d x} [- 8 e^{- 8 x}]$ .

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Step 4.1.2.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 e^{- 8 x}$ with respect to $x$ is $- 8 \frac{d}{d x} [e^{- 8 x}]$ .

$- 8 \frac{d}{d x} [e^{- 8 x}] + \frac{d}{d x} [5 e^{- 5 x}]$

Step 4.1.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 8 x$ .

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Step 4.1.2.2.1

To apply the Chain Rule, set $u_{1}$ as $- 8 x$ .

$- 8 (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [- 8 x]) + \frac{d}{d x} [5 e^{- 5 x}]$

Step 4.1.2.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$- 8 (e^{u_{1}} \frac{d}{d x} [- 8 x]) + \frac{d}{d x} [5 e^{- 5 x}]$

Step 4.1.2.2.3

Replace all occurrences of $u_{1}$ with $- 8 x$ .

$- 8 (e^{- 8 x} \frac{d}{d x} [- 8 x]) + \frac{d}{d x} [5 e^{- 5 x}]$

Step 4.1.2.3

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 x$ with respect to $x$ is $- 8 \frac{d}{d x} [x]$ .

$- 8 (e^{- 8 x} (- 8 \frac{d}{d x} [x])) + \frac{d}{d x} [5 e^{- 5 x}]$

Step 4.1.2.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$- 8 (e^{- 8 x} (- 8 \cdot 1)) + \frac{d}{d x} [5 e^{- 5 x}]$

Step 4.1.2.5

Multiply $- 8$ by $1$ .

$- 8 (e^{- 8 x} \cdot - 8) + \frac{d}{d x} [5 e^{- 5 x}]$

Step 4.1.2.6

Move $- 8$ to the left of $e^{- 8 x}$ .

$- 8 (- 8 \cdot e^{- 8 x}) + \frac{d}{d x} [5 e^{- 5 x}]$

Step 4.1.2.7

Multiply $- 8$ by $- 8$ .

$64 e^{- 8 x} + \frac{d}{d x} [5 e^{- 5 x}]$

Step 4.1.3

Evaluate $\frac{d}{d x} [5 e^{- 5 x}]$ .

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Step 4.1.3.1

Since $5$ is constant with respect to $x$ , the derivative of $5 e^{- 5 x}$ with respect to $x$ is $5 \frac{d}{d x} [e^{- 5 x}]$ .

$64 e^{- 8 x} + 5 \frac{d}{d x} [e^{- 5 x}]$

Step 4.1.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 5 x$ .

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Step 4.1.3.2.1

To apply the Chain Rule, set $u_{2}$ as $- 5 x$ .

$64 e^{- 8 x} + 5 (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- 5 x])$

Step 4.1.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$64 e^{- 8 x} + 5 (e^{u_{2}} \frac{d}{d x} [- 5 x])$

Step 4.1.3.2.3

Replace all occurrences of $u_{2}$ with $- 5 x$ .

$64 e^{- 8 x} + 5 (e^{- 5 x} \frac{d}{d x} [- 5 x])$

Step 4.1.3.3

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x$ with respect to $x$ is $- 5 \frac{d}{d x} [x]$ .

$64 e^{- 8 x} + 5 (e^{- 5 x} (- 5 \frac{d}{d x} [x]))$

Step 4.1.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$64 e^{- 8 x} + 5 (e^{- 5 x} (- 5 \cdot 1))$

Step 4.1.3.5

Multiply $- 5$ by $1$ .

$64 e^{- 8 x} + 5 (e^{- 5 x} \cdot - 5)$

Step 4.1.3.6

Move $- 5$ to the left of $e^{- 5 x}$ .

$64 e^{- 8 x} + 5 (- 5 \cdot e^{- 5 x})$

Step 4.1.3.7

Multiply $- 5$ by $5$ .

$f' (x) = 64 e^{- 8 x} - 25 e^{- 5 x}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $64 e^{- 8 x} - 25 e^{- 5 x}$ .

$64 e^{- 8 x} - 25 e^{- 5 x}$

Step 5

Set the first derivative equal to $0$ then solve the equation $64 e^{- 8 x} - 25 e^{- 5 x} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$64 e^{- 8 x} - 25 e^{- 5 x} = 0$

Step 5.2

Move $- 25 e^{- 5 x}$ to the right side of the equation by adding it to both sides.

$64 e^{- 8 x} = 25 e^{- 5 x}$

Step 5.3

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (64 e^{- 8 x}) = \ln (25 e^{- 5 x})$

Step 5.4

Expand the left side.

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Step 5.4.1

Rewrite $\ln (64 e^{- 8 x})$ as $\ln (64) + \ln (e^{- 8 x})$ .

$\ln (64) + \ln (e^{- 8 x}) = \ln (25 e^{- 5 x})$

Step 5.4.2

Expand $\ln (e^{- 8 x})$ by moving $- 8 x$ outside the logarithm.

$\ln (64) - 8 x \ln (e) = \ln (25 e^{- 5 x})$

Step 5.4.3

The natural logarithm of $e$ is $1$ .

$\ln (64) - 8 x \cdot 1 = \ln (25 e^{- 5 x})$

Step 5.4.4

Multiply $- 8$ by $1$ .

$\ln (64) - 8 x = \ln (25 e^{- 5 x})$

Step 5.5

Expand the right side.

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Step 5.5.1

Rewrite $\ln (25 e^{- 5 x})$ as $\ln (25) + \ln (e^{- 5 x})$ .

$\ln (64) - 8 x = \ln (25) + \ln (e^{- 5 x})$

Step 5.5.2

Expand $\ln (e^{- 5 x})$ by moving $- 5 x$ outside the logarithm.

$\ln (64) - 8 x = \ln (25) - 5 x \ln (e)$

Step 5.5.3

The natural logarithm of $e$ is $1$ .

$\ln (64) - 8 x = \ln (25) - 5 x \cdot 1$

Step 5.5.4

Multiply $- 5$ by $1$ .

$\ln (64) - 8 x = \ln (25) - 5 x$

Step 5.6

Move all the terms containing a logarithm to the left side of the equation.

$\ln (64) - \ln (25) = 8 x - 5 x$

Step 5.7

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{64}{25}) = 8 x - 5 x$

Step 5.8

Subtract $5 x$ from $8 x$ .

$\ln (\frac{64}{25}) = 3 x$

Step 5.9

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$3 x = \ln (\frac{64}{25})$

Step 5.10

Divide each term in $3 x = \ln (\frac{64}{25})$ by $3$ and simplify.

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Step 5.10.1

Divide each term in $3 x = \ln (\frac{64}{25})$ by $3$ .

$\frac{3 x}{3} = \frac{\ln (\frac{64}{25})}{3}$

Step 5.10.2

Simplify the left side.

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Step 5.10.2.1

Cancel the common factor of $3$ .

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Step 5.10.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{\ln (\frac{64}{25})}{3}$

Step 5.10.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (\frac{64}{25})}{3}$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$x = \frac{\ln (\frac{64}{25})}{3}$

Step 8

Evaluate the second derivative at $x = \frac{\ln (\frac{64}{25})}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$125 e^{- 5 \frac{\ln (\frac{64}{25})}{3}} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9

Simplify each term.

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Step 9.1

Rewrite $\frac{\ln (\frac{64}{25})}{3}$ as $\frac{1}{3} \ln (\frac{64}{25})$ .

$125 e^{- 5 (\frac{1}{3} \ln (\frac{64}{25}))} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.2

Simplify $\frac{1}{3} \ln (\frac{64}{25})$ by moving $\frac{1}{3}$ inside the logarithm.

$125 e^{- 5 \ln ({(\frac{64}{25})}^{\frac{1}{3}})} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.3

Simplify $- 5 \ln ({(\frac{64}{25})}^{\frac{1}{3}})$ by moving $- 5$ inside the logarithm.

$125 e^{\ln ({({(\frac{64}{25})}^{\frac{1}{3}})}^{- 5})} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.4

Exponentiation and log are inverse functions.

$125 {({(\frac{64}{25})}^{\frac{1}{3}})}^{- 5} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.5

Multiply the exponents in ${({(\frac{64}{25})}^{\frac{1}{3}})}^{- 5}$ .

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Step 9.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$125 {(\frac{64}{25})}^{\frac{1}{3} \cdot - 5} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.5.2

Combine $\frac{1}{3}$ and $- 5$ .

$125 {(\frac{64}{25})}^{\frac{- 5}{3}} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.5.3

Move the negative in front of the fraction.

$125 {(\frac{64}{25})}^{- \frac{5}{3}} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.6

Change the sign of the exponent by rewriting the base as its reciprocal.

$125 {(\frac{25}{64})}^{\frac{5}{3}} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.7

Apply the product rule to $\frac{25}{64}$ .

$125 \frac{25^{\frac{5}{3}}}{64^{\frac{5}{3}}} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.8

Simplify the denominator.

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Step 9.8.1

Rewrite $64$ as $4^{3}$ .

$125 \frac{25^{\frac{5}{3}}}{{(4^{3})}^{\frac{5}{3}}} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.8.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$125 \frac{25^{\frac{5}{3}}}{4^{3 (\frac{5}{3})}} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.8.3

Cancel the common factor of $3$ .

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Step 9.8.3.1

Cancel the common factor.

$125 \frac{25^{\frac{5}{3}}}{4^{3 (\frac{5}{3})}} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.8.3.2

Rewrite the expression.

$125 \frac{25^{\frac{5}{3}}}{4^{5}} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.8.4

Raise $4$ to the power of $5$ .

$125 \frac{25^{\frac{5}{3}}}{1024} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.9

Multiply $125 \frac{25^{\frac{5}{3}}}{1024}$ .

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Step 9.9.1

Combine $125$ and $\frac{25^{\frac{5}{3}}}{1024}$ .

$\frac{125 \cdot 25^{\frac{5}{3}}}{1024} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.9.2

Rewrite $125$ as $5^{3}$ .

$\frac{5^{3} \cdot 25^{\frac{5}{3}}}{1024} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.9.3

Rewrite $25$ as $5^{2}$ .

$\frac{5^{3} \cdot {(5^{2})}^{\frac{5}{3}}}{1024} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.9.4

Multiply the exponents in ${(5^{2})}^{\frac{5}{3}}$ .

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Step 9.9.4.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{5^{3} \cdot 5^{2 (\frac{5}{3})}}{1024} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.9.4.2

Multiply $2 (\frac{5}{3})$ .

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Step 9.9.4.2.1

Combine $2$ and $\frac{5}{3}$ .

$\frac{5^{3} \cdot 5^{\frac{2 \cdot 5}{3}}}{1024} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.9.4.2.2

Multiply $2$ by $5$ .

$\frac{5^{3} \cdot 5^{\frac{10}{3}}}{1024} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.9.5

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{5^{3 + \frac{10}{3}}}{1024} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.9.6

To write $3$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{5^{3 \cdot \frac{3}{3} + \frac{10}{3}}}{1024} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.9.7

Combine $3$ and $\frac{3}{3}$ .

$\frac{5^{\frac{3 \cdot 3}{3} + \frac{10}{3}}}{1024} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.9.8

Combine the numerators over the common denominator.

$\frac{5^{\frac{3 \cdot 3 + 10}{3}}}{1024} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.9.9

Simplify the numerator.

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Step 9.9.9.1

Multiply $3$ by $3$ .

$\frac{5^{\frac{9 + 10}{3}}}{1024} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.9.9.2

Add $9$ and $10$ .

$\frac{5^{\frac{19}{3}}}{1024} - 512 e^{- 8 \frac{\ln (\frac{64}{25})}{3}}$

Step 9.10

Rewrite $\frac{\ln (\frac{64}{25})}{3}$ as $\frac{1}{3} \ln (\frac{64}{25})$ .

$\frac{5^{\frac{19}{3}}}{1024} - 512 e^{- 8 (\frac{1}{3} \ln (\frac{64}{25}))}$

Step 9.11

Simplify $\frac{1}{3} \ln (\frac{64}{25})$ by moving $\frac{1}{3}$ inside the logarithm.

$\frac{5^{\frac{19}{3}}}{1024} - 512 e^{- 8 \ln ({(\frac{64}{25})}^{\frac{1}{3}})}$

Step 9.12

Simplify $- 8 \ln ({(\frac{64}{25})}^{\frac{1}{3}})$ by moving $- 8$ inside the logarithm.

$\frac{5^{\frac{19}{3}}}{1024} - 512 e^{\ln ({({(\frac{64}{25})}^{\frac{1}{3}})}^{- 8})}$

Step 9.13

Exponentiation and log are inverse functions.

$\frac{5^{\frac{19}{3}}}{1024} - 512 {({(\frac{64}{25})}^{\frac{1}{3}})}^{- 8}$

Step 9.14

Multiply the exponents in ${({(\frac{64}{25})}^{\frac{1}{3}})}^{- 8}$ .

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Step 9.14.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{5^{\frac{19}{3}}}{1024} - 512 {(\frac{64}{25})}^{\frac{1}{3} \cdot - 8}$

Step 9.14.2

Combine $\frac{1}{3}$ and $- 8$ .

$\frac{5^{\frac{19}{3}}}{1024} - 512 {(\frac{64}{25})}^{\frac{- 8}{3}}$

Step 9.14.3

Move the negative in front of the fraction.

$\frac{5^{\frac{19}{3}}}{1024} - 512 {(\frac{64}{25})}^{- \frac{8}{3}}$

Step 9.15

Change the sign of the exponent by rewriting the base as its reciprocal.

$\frac{5^{\frac{19}{3}}}{1024} - 512 {(\frac{25}{64})}^{\frac{8}{3}}$

Step 9.16

Apply the product rule to $\frac{25}{64}$ .

$\frac{5^{\frac{19}{3}}}{1024} - 512 \frac{25^{\frac{8}{3}}}{64^{\frac{8}{3}}}$

Step 9.17

Simplify the denominator.

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Step 9.17.1

Rewrite $64$ as $4^{3}$ .

$\frac{5^{\frac{19}{3}}}{1024} - 512 \frac{25^{\frac{8}{3}}}{{(4^{3})}^{\frac{8}{3}}}$

Step 9.17.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{5^{\frac{19}{3}}}{1024} - 512 \frac{25^{\frac{8}{3}}}{4^{3 (\frac{8}{3})}}$

Step 9.17.3

Cancel the common factor of $3$ .

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Step 9.17.3.1

Cancel the common factor.

$\frac{5^{\frac{19}{3}}}{1024} - 512 \frac{25^{\frac{8}{3}}}{4^{3 (\frac{8}{3})}}$

Step 9.17.3.2

Rewrite the expression.

$\frac{5^{\frac{19}{3}}}{1024} - 512 \frac{25^{\frac{8}{3}}}{4^{8}}$

Step 9.17.4

Raise $4$ to the power of $8$ .

$\frac{5^{\frac{19}{3}}}{1024} - 512 \frac{25^{\frac{8}{3}}}{65536}$

Step 9.18

Cancel the common factor of $512$ .

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Step 9.18.1

Factor $512$ out of $- 512$ .

$\frac{5^{\frac{19}{3}}}{1024} + 512 (- 1) \frac{25^{\frac{8}{3}}}{65536}$

Step 9.18.2

Factor $512$ out of $65536$ .

$\frac{5^{\frac{19}{3}}}{1024} + 512 \cdot - 1 \frac{25^{\frac{8}{3}}}{512 \cdot 128}$

Step 9.18.3

Cancel the common factor.

$\frac{5^{\frac{19}{3}}}{1024} + 512 \cdot - 1 \frac{25^{\frac{8}{3}}}{512 \cdot 128}$

Step 9.18.4

Rewrite the expression.

$\frac{5^{\frac{19}{3}}}{1024} - 1 \frac{25^{\frac{8}{3}}}{128}$

Step 9.19

Rewrite $- 1 \frac{25^{\frac{8}{3}}}{128}$ as $- \frac{25^{\frac{8}{3}}}{128}$ .

$\frac{5^{\frac{19}{3}}}{1024} - \frac{25^{\frac{8}{3}}}{128}$

Step 10

$x = \frac{\ln (\frac{64}{25})}{3}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{\ln (\frac{64}{25})}{3}$ is a local maximum

Step 11

Find the y-value when $x = \frac{\ln (\frac{64}{25})}{3}$ .

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Step 11.1

Simplify $\frac{\ln (\frac{64}{25})}{3}$ to substitute in $f (x) = - 8 e^{- 8 x} + 5 e^{- 5 x}$ .

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Step 11.1.1

Rewrite $\frac{\ln (\frac{64}{25})}{3}$ as $\frac{1}{3} \ln (\frac{64}{25})$ .

$y = \frac{1}{3} \cdot \ln (\frac{64}{25})$

Step 11.1.2

Simplify $\frac{1}{3} \ln (\frac{64}{25})$ by moving $\frac{1}{3}$ inside the logarithm.

$y = \ln ({(\frac{64}{25})}^{\frac{1}{3}})$

Step 11.1.3

Apply the product rule to $\frac{64}{25}$ .

$y = \ln (\frac{64^{\frac{1}{3}}}{25^{\frac{1}{3}}})$

Step 11.1.4

Simplify the numerator.

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Step 11.1.4.1

Rewrite $64$ as $4^{3}$ .

$y = \ln (\frac{{(4^{3})}^{\frac{1}{3}}}{25^{\frac{1}{3}}})$

Step 11.1.4.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y = \ln (\frac{4^{3 (\frac{1}{3})}}{25^{\frac{1}{3}}})$

Step 11.1.4.3

Cancel the common factor of $3$ .

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Step 11.1.4.3.1

Cancel the common factor.

$y = \ln (\frac{4^{3 (\frac{1}{3})}}{25^{\frac{1}{3}}})$

Step 11.1.4.3.2

Rewrite the expression.

$y = \ln (\frac{4}{25^{\frac{1}{3}}})$

Step 11.1.4.4

Evaluate the exponent.

$y = \ln (\frac{4}{25^{\frac{1}{3}}})$

Step 11.2

Replace the variable $x$ with $\ln (\frac{4}{25^{\frac{1}{3}}})$ in the expression.

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - 8 e^{- 8 \ln (\frac{4}{25^{\frac{1}{3}}})} + 5 e^{- 5 \ln (\frac{4}{25^{\frac{1}{3}}})}$

Step 11.3

Simplify the result.

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Step 11.3.1

Simplify each term.

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Step 11.3.1.1

Simplify $- 8 \ln (\frac{4}{25^{\frac{1}{3}}})$ by moving $- 8$ inside the logarithm.

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - 8 e^{\ln ({(\frac{4}{25^{\frac{1}{3}}})}^{- 8})} + 5 e^{- 5 \ln (\frac{4}{25^{\frac{1}{3}}})}$

Step 11.3.1.2

Exponentiation and log are inverse functions.

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - 8 {(\frac{4}{25^{\frac{1}{3}}})}^{- 8} + 5 e^{- 5 \ln (\frac{4}{25^{\frac{1}{3}}})}$

Step 11.3.1.3

Change the sign of the exponent by rewriting the base as its reciprocal.

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - 8 {(\frac{25^{\frac{1}{3}}}{4})}^{8} + 5 e^{- 5 \ln (\frac{4}{25^{\frac{1}{3}}})}$

Step 11.3.1.4

Apply the product rule to $\frac{25^{\frac{1}{3}}}{4}$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - 8 \frac{{(25^{\frac{1}{3}})}^{8}}{4^{8}} + 5 e^{- 5 \ln (\frac{4}{25^{\frac{1}{3}}})}$

Step 11.3.1.5

Multiply the exponents in ${(25^{\frac{1}{3}})}^{8}$ .

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Step 11.3.1.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - 8 \frac{25^{\frac{1}{3} \cdot 8}}{4^{8}} + 5 e^{- 5 \ln (\frac{4}{25^{\frac{1}{3}}})}$

Step 11.3.1.5.2

Combine $\frac{1}{3}$ and $8$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - 8 \frac{25^{\frac{8}{3}}}{4^{8}} + 5 e^{- 5 \ln (\frac{4}{25^{\frac{1}{3}}})}$

Step 11.3.1.6

Raise $4$ to the power of $8$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - 8 \frac{25^{\frac{8}{3}}}{65536} + 5 e^{- 5 \ln (\frac{4}{25^{\frac{1}{3}}})}$

Step 11.3.1.7

Cancel the common factor of $8$ .

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Step 11.3.1.7.1

Factor $8$ out of $- 8$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = 8 (- 1) (\frac{25^{\frac{8}{3}}}{65536}) + 5 e^{- 5 \ln (\frac{4}{25^{\frac{1}{3}}})}$

Step 11.3.1.7.2

Factor $8$ out of $65536$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = 8 \cdot (- 1 \frac{25^{\frac{8}{3}}}{8 \cdot 8192}) + 5 e^{- 5 \ln (\frac{4}{25^{\frac{1}{3}}})}$

Step 11.3.1.7.3

Cancel the common factor.

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = 8 \cdot (- 1 \frac{25^{\frac{8}{3}}}{8 \cdot 8192}) + 5 e^{- 5 \ln (\frac{4}{25^{\frac{1}{3}}})}$

Step 11.3.1.7.4

Rewrite the expression.

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - 1 \frac{25^{\frac{8}{3}}}{8192} + 5 e^{- 5 \ln (\frac{4}{25^{\frac{1}{3}}})}$

Step 11.3.1.8

Rewrite $- 1 \frac{25^{\frac{8}{3}}}{8192}$ as $- \frac{25^{\frac{8}{3}}}{8192}$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + 5 e^{- 5 \ln (\frac{4}{25^{\frac{1}{3}}})}$

Step 11.3.1.9

Simplify $- 5 \ln (\frac{4}{25^{\frac{1}{3}}})$ by moving $- 5$ inside the logarithm.

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + 5 e^{\ln ({(\frac{4}{25^{\frac{1}{3}}})}^{- 5})}$

Step 11.3.1.10

Exponentiation and log are inverse functions.

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + 5 {(\frac{4}{25^{\frac{1}{3}}})}^{- 5}$

Step 11.3.1.11

Change the sign of the exponent by rewriting the base as its reciprocal.

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + 5 {(\frac{25^{\frac{1}{3}}}{4})}^{5}$

Step 11.3.1.12

Apply the product rule to $\frac{25^{\frac{1}{3}}}{4}$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + 5 (\frac{{(25^{\frac{1}{3}})}^{5}}{4^{5}})$

Step 11.3.1.13

Multiply the exponents in ${(25^{\frac{1}{3}})}^{5}$ .

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Step 11.3.1.13.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + 5 (\frac{25^{\frac{1}{3} \cdot 5}}{4^{5}})$

Step 11.3.1.13.2

Combine $\frac{1}{3}$ and $5$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + 5 (\frac{25^{\frac{5}{3}}}{4^{5}})$

Step 11.3.1.14

Raise $4$ to the power of $5$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + 5 (\frac{25^{\frac{5}{3}}}{1024})$

Step 11.3.1.15

Multiply $5 \frac{25^{\frac{5}{3}}}{1024}$ .

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Step 11.3.1.15.1

Combine $5$ and $\frac{25^{\frac{5}{3}}}{1024}$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + \frac{5 \cdot 25^{\frac{5}{3}}}{1024}$

Step 11.3.1.15.2

Rewrite $25$ as $5^{2}$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + \frac{5 \cdot {(5^{2})}^{\frac{5}{3}}}{1024}$

Step 11.3.1.15.3

Multiply the exponents in ${(5^{2})}^{\frac{5}{3}}$ .

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Step 11.3.1.15.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + \frac{5 \cdot 5^{2 (\frac{5}{3})}}{1024}$

Step 11.3.1.15.3.2

Multiply $2 (\frac{5}{3})$ .

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Step 11.3.1.15.3.2.1

Combine $2$ and $\frac{5}{3}$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + \frac{5 \cdot 5^{\frac{2 \cdot 5}{3}}}{1024}$

Step 11.3.1.15.3.2.2

Multiply $2$ by $5$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + \frac{5 \cdot 5^{\frac{10}{3}}}{1024}$

Step 11.3.1.15.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + \frac{5^{1 + \frac{10}{3}}}{1024}$

Step 11.3.1.15.5

Write $1$ as a fraction with a common denominator.

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + \frac{5^{\frac{3}{3} + \frac{10}{3}}}{1024}$

Step 11.3.1.15.6

Combine the numerators over the common denominator.

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + \frac{5^{\frac{3 + 10}{3}}}{1024}$

Step 11.3.1.15.7

Add $3$ and $10$ .

$f (\ln (\frac{4}{25^{\frac{1}{3}}})) = - \frac{25^{\frac{8}{3}}}{8192} + \frac{5^{\frac{13}{3}}}{1024}$

Step 11.3.2

The final answer is $- \frac{25^{\frac{8}{3}}}{8192} + \frac{5^{\frac{13}{3}}}{1024}$ .

$y = - \frac{25^{\frac{8}{3}}}{8192} + \frac{5^{\frac{13}{3}}}{1024}$

Step 12

These are the local extrema for $f (x) = - 8 e^{- 8 x} + 5 e^{- 5 x}$ .

$(\frac{\ln (\frac{64}{25})}{3}, - \frac{25^{\frac{8}{3}}}{8192} + \frac{5^{\frac{13}{3}}}{1024})$ is a local maxima

Step 13