Algebra Examples

$4^{3 x + 2} = (x - 5) \cdot 8^{2 x}$

Step 1

Take the log of both sides of the equation.

$\ln (4^{3 x + 2}) = \ln ((x - 5) \cdot 8^{2 x})$

Step 2

Expand $\ln (4^{3 x + 2})$ by moving $3 x + 2$ outside the logarithm.

$(3 x + 2) \ln (4) = \ln ((x - 5) \cdot 8^{2 x})$

Step 3

Rewrite $\ln ((x - 5) \cdot 8^{2 x})$ as $\ln (x - 5) + \ln (8^{2 x})$ .

$(3 x + 2) \ln (4) = \ln (x - 5) + \ln (8^{2 x})$

Step 4

Expand $\ln (8^{2 x})$ by moving $2 x$ outside the logarithm.

$(3 x + 2) \ln (4) = \ln (x - 5) + 2 x \ln (8)$

Step 5

Solve the equation for $x$ .

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Step 5.1

Simplify the left side.

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Step 5.1.1

Simplify $(3 x + 2) \ln (4)$ .

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Step 5.1.1.1

Apply the distributive property.

$3 x \ln (4) + 2 \ln (4) = \ln (x - 5) + 2 x \ln (8)$

Step 5.1.1.2

Simplify $3 \ln (4)$ by moving $3$ inside the logarithm.

$x \ln (4^{3}) + 2 \ln (4) = \ln (x - 5) + 2 x \ln (8)$

Step 5.1.1.3

Simplify $2 \ln (4)$ by moving $2$ inside the logarithm.

$x \ln (4^{3}) + \ln (4^{2}) = \ln (x - 5) + 2 x \ln (8)$

Step 5.1.1.4

Simplify each term.

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Step 5.1.1.4.1

Raise $4$ to the power of $3$ .

$x \ln (64) + \ln (4^{2}) = \ln (x - 5) + 2 x \ln (8)$

Step 5.1.1.4.2

Raise $4$ to the power of $2$ .

$x \ln (64) + \ln (16) = \ln (x - 5) + 2 x \ln (8)$

Step 5.2

Simplify the right side.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Simplify $2 \ln (8)$ by moving $2$ inside the logarithm.

$x \ln (64) + \ln (16) = \ln (x - 5) + x \ln (8^{2})$

Step 5.2.1.2

Raise $8$ to the power of $2$ .

$x \ln (64) + \ln (16) = \ln (x - 5) + x \ln (64)$

Step 5.3

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

$x \ln (64) + \ln (16) - \ln (x - 5) - x \ln (64) = 0$

Step 5.4

Combine the opposite terms in $x \ln (64) + \ln (16) - \ln (x - 5) - x \ln (64)$ .

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Step 5.4.1

Subtract $x \ln (64)$ from $x \ln (64)$ .

$0 + \ln (16) - \ln (x - 5) = 0$

Step 5.4.2

Add $0$ and $\ln (16)$ .

$\ln (16) - \ln (x - 5) = 0$

Step 5.5

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

$\ln (\frac{16}{x - 5}) = 0$

Step 5.6

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (\frac{16}{x - 5})} = e^{0}$

Step 5.7

Rewrite $\ln (\frac{16}{x - 5}) = 0$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{0} = \frac{16}{x - 5}$

Step 5.8

Solve for $x$ .

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Step 5.8.1

Rewrite the equation as $\frac{16}{x - 5} = e^{0}$ .

$\frac{16}{x - 5} = e^{0}$

Step 5.8.2

Anything raised to $0$ is $1$ .

$\frac{16}{x - 5} = 1$

Step 5.8.3

Find the LCD of the terms in the equation.

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Step 5.8.3.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x - 5, 1$

Step 5.8.3.2

Remove parentheses.

$x - 5, 1$

Step 5.8.3.3

The LCM of one and any expression is the expression.

$x - 5$

Step 5.8.4

Multiply each term in $\frac{16}{x - 5} = 1$ by $x - 5$ to eliminate the fractions.

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Step 5.8.4.1

Multiply each term in $\frac{16}{x - 5} = 1$ by $x - 5$ .

$\frac{16}{x - 5} (x - 5) = 1 (x - 5)$

Step 5.8.4.2

Simplify the left side.

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Step 5.8.4.2.1

Cancel the common factor of $x - 5$ .

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Step 5.8.4.2.1.1

Cancel the common factor.

$\frac{16}{x - 5} (x - 5) = 1 (x - 5)$

Step 5.8.4.2.1.2

Rewrite the expression.

$16 = 1 (x - 5)$

Step 5.8.4.3

Simplify the right side.

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Step 5.8.4.3.1

Multiply $x - 5$ by $1$ .

$16 = x - 5$

Step 5.8.5

Solve the equation.

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Step 5.8.5.1

Rewrite the equation as $x - 5 = 16$ .

$x - 5 = 16$

Step 5.8.5.2

Move all terms not containing $x$ to the right side of the equation.

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Step 5.8.5.2.1

Add $5$ to both sides of the equation.

$x = 16 + 5$

Step 5.8.5.2.2

Add $16$ and $5$ .

$x = 21$