Algebra Examples

$f (x) = {(\frac{3}{2})}^{- x} - 7$

Step 1

Find the x-intercepts.

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Step 1.1

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = {(\frac{3}{2})}^{- x} - 7$

Step 1.2

Solve the equation.

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Step 1.2.1

Rewrite the equation as ${(\frac{3}{2})}^{- x} - 7 = 0$ .

${(\frac{3}{2})}^{- x} - 7 = 0$

Step 1.2.2

Apply the product rule to $\frac{3}{2}$ .

$\frac{3^{- x}}{2^{- x}} - 7 = 0$

Step 1.2.3

Add $7$ to both sides of the equation.

$\frac{3^{- x}}{2^{- x}} = 7$

Step 1.2.4

Multiply both sides by $2^{- x}$ .

$\frac{3^{- x}}{2^{- x}} \cdot 2^{- x} = 7 \cdot 2^{- x}$

Step 1.2.5

Simplify the left side.

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Step 1.2.5.1

Cancel the common factor of $2^{- x}$ .

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Step 1.2.5.1.1

Cancel the common factor.

$\frac{3^{- x}}{2^{- x}} \cdot 2^{- x} = 7 \cdot 2^{- x}$

Step 1.2.5.1.2

Rewrite the expression.

$3^{- x} = 7 \cdot 2^{- x}$

Step 1.2.6

Solve for $x$ .

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Step 1.2.6.1

Take the log of both sides of the equation.

$\ln (3^{- x}) = \ln (7 \cdot 2^{- x})$

Step 1.2.6.2

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

$- x \ln (3) = \ln (7 \cdot 2^{- x})$

Step 1.2.6.3

Rewrite $\ln (7 \cdot 2^{- x})$ as $\ln (7) + \ln (2^{- x})$ .

$- x \ln (3) = \ln (7) + \ln (2^{- x})$

Step 1.2.6.4

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

$- x \ln (3) = \ln (7) - x \ln (2)$

Step 1.2.6.5

Solve the equation for $x$ .

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Step 1.2.6.5.1

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

$- x \ln (3) - \ln (7) + x \ln (2) = 0$

Step 1.2.6.5.2

Add $\ln (7)$ to both sides of the equation.

$- x \ln (3) + x \ln (2) = \ln (7)$

Step 1.2.6.5.3

Factor $x$ out of $- x \ln (3) + x \ln (2)$ .

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Step 1.2.6.5.3.1

Factor $x$ out of $- x \ln (3)$ .

$x (- 1 \ln (3)) + x \ln (2) = \ln (7)$

Step 1.2.6.5.3.2

Factor $x$ out of $x \ln (2)$ .

$x (- 1 \ln (3)) + x (\ln (2)) = \ln (7)$

Step 1.2.6.5.3.3

Factor $x$ out of $x (- 1 \ln (3)) + x (\ln (2))$ .

$x (- 1 \ln (3) + \ln (2)) = \ln (7)$

Step 1.2.6.5.4

Rewrite $- 1 \ln (3)$ as $- \ln (3)$ .

$x (- \ln (3) + \ln (2)) = \ln (7)$

Step 1.2.6.5.5

Divide each term in $x (- \ln (3) + \ln (2)) = \ln (7)$ by $- \ln (3) + \ln (2)$ and simplify.

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Step 1.2.6.5.5.1

Divide each term in $x (- \ln (3) + \ln (2)) = \ln (7)$ by $- \ln (3) + \ln (2)$ .

$\frac{x (- \ln (3) + \ln (2))}{- \ln (3) + \ln (2)} = \frac{\ln (7)}{- \ln (3) + \ln (2)}$

Step 1.2.6.5.5.2

Simplify the left side.

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Step 1.2.6.5.5.2.1

Cancel the common factor of $- \ln (3) + \ln (2)$ .

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Step 1.2.6.5.5.2.1.1

Cancel the common factor.

$\frac{x (- \ln (3) + \ln (2))}{- \ln (3) + \ln (2)} = \frac{\ln (7)}{- \ln (3) + \ln (2)}$

Step 1.2.6.5.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (7)}{- \ln (3) + \ln (2)}$

Step 1.2.6.5.5.3

Simplify the right side.

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Step 1.2.6.5.5.3.1

Factor $- 1$ out of $- \ln (3)$ .

$x = \frac{\ln (7)}{- \ln (3) + \ln (2)}$

Step 1.2.6.5.5.3.2

Factor $- 1$ out of $\ln (2)$ .

$x = \frac{\ln (7)}{- \ln (3) - 1 (- \ln (2))}$

Step 1.2.6.5.5.3.3

Factor $- 1$ out of $- \ln (3) - 1 (- \ln (2))$ .

$x = \frac{\ln (7)}{- (\ln (3) - \ln (2))}$

Step 1.2.6.5.5.3.4

Rewrite negatives.

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Step 1.2.6.5.5.3.4.1

Rewrite $- (\ln (3) - \ln (2))$ as $- 1 (\ln (3) - \ln (2))$ .

$x = \frac{\ln (7)}{- 1 (\ln (3) - \ln (2))}$

Step 1.2.6.5.5.3.4.2

Move the negative in front of the fraction.

$x = - \frac{\ln (7)}{\ln (3) - \ln (2)}$

Step 1.3

x-intercept(s) in point form.

x-intercept(s): $(- \frac{\ln (7)}{\ln (3) - \ln (2)}, 0)$

Step 2

Find the y-intercepts.

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Step 2.1

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = {(\frac{3}{2})}^{- (0)} - 7$

Step 2.2

Solve the equation.

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Step 2.2.1

Remove parentheses.

$y = {(\frac{3}{2})}^{- (0)} - 7$

Step 2.2.2

Simplify ${(\frac{3}{2})}^{- (0)} - 7$ .

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

Multiply $- 1$ by $0$ .

$y = {(\frac{3}{2})}^{0} - 7$

Step 2.2.2.1.2

Apply the product rule to $\frac{3}{2}$ .

$y = \frac{3^{0}}{2^{0}} - 7$

Step 2.2.2.1.3

Anything raised to $0$ is $1$ .

$y = \frac{1}{2^{0}} - 7$

Step 2.2.2.1.4

Anything raised to $0$ is $1$ .

$y = \frac{1}{1} - 7$

Step 2.2.2.1.5

Divide $1$ by $1$ .

$y = 1 - 7$

Step 2.2.2.2

Subtract $7$ from $1$ .

$y = - 6$

Step 2.3

y-intercept(s) in point form.

y-intercept(s): $(0, - 6)$

Step 3

List the intersections.

x-intercept(s): $(- \frac{\ln (7)}{\ln (3) - \ln (2)}, 0)$

y-intercept(s): $(0, - 6)$

Step 4