Algebra Examples

$f (x) = {(\frac{1}{2})}^{- x} - 5$

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 $x$ .

$0 = {(\frac{1}{2})}^{- x} - 5$

Step 1.2

Solve the equation.

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Step 1.2.1

Rewrite the equation as ${(\frac{1}{2})}^{- x} - 5 = 0$ .

${(\frac{1}{2})}^{- x} - 5 = 0$

Step 1.2.2

Simplify each term.

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Step 1.2.2.1

Apply the product rule to $\frac{1}{2}$ .

$\frac{1^{- x}}{2^{- x}} - 5 = 0$

Step 1.2.2.2

One to any power is one.

$\frac{1}{2^{- x}} - 5 = 0$

Step 1.2.3

Add $5$ to both sides of the equation.

$\frac{1}{2^{- x}} = 5$

Step 1.2.4

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

$\frac{1}{2^{- x}} \cdot 2^{- x} = 5 \cdot 2^{- x}$

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.

$\frac{1}{2^{- x}} \cdot 2^{- x} = 5 \cdot 2^{- x}$

Step 1.2.5.1.2

Rewrite the expression.

$1 = 5 \cdot 2^{- x}$

Step 1.2.6

Solve for $x$ .

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Step 1.2.6.1

Rewrite the equation as $5 \cdot 2^{- x} = 1$ .

$5 \cdot 2^{- x} = 1$

Step 1.2.6.2

Divide each term in $5 \cdot 2^{- x} = 1$ by $5$ and simplify.

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Step 1.2.6.2.1

Divide each term in $5 \cdot 2^{- x} = 1$ by $5$ .

$\frac{5 \cdot 2^{- x}}{5} = \frac{1}{5}$

Step 1.2.6.2.2

Simplify the left side.

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Step 1.2.6.2.2.1

Cancel the common factor of $5$ .

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Step 1.2.6.2.2.1.1

Cancel the common factor.

$\frac{5 \cdot 2^{- x}}{5} = \frac{1}{5}$

Step 1.2.6.2.2.1.2

Divide $2^{- x}$ by $1$ .

$2^{- x} = \frac{1}{5}$

Step 1.2.6.3

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

$\ln (2^{- x}) = \ln (\frac{1}{5})$

Step 1.2.6.4

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

$- x \ln (2) = \ln (\frac{1}{5})$

Step 1.2.6.5

Divide each term in $- x \ln (2) = \ln (\frac{1}{5})$ by $- \ln (2)$ and simplify.

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Step 1.2.6.5.1

Divide each term in $- x \ln (2) = \ln (\frac{1}{5})$ by $- \ln (2)$ .

$\frac{- x \ln (2)}{- \ln (2)} = \frac{\ln (\frac{1}{5})}{- \ln (2)}$

Step 1.2.6.5.2

Simplify the left side.

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Step 1.2.6.5.2.1

Dividing two negative values results in a positive value.

$\frac{x \ln (2)}{\ln (2)} = \frac{\ln (\frac{1}{5})}{- \ln (2)}$

Step 1.2.6.5.2.2

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

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Step 1.2.6.5.2.2.1

Cancel the common factor.

$\frac{x \ln (2)}{\ln (2)} = \frac{\ln (\frac{1}{5})}{- \ln (2)}$

Step 1.2.6.5.2.2.2

Divide $x$ by $1$ .

$x = \frac{\ln (\frac{1}{5})}{- \ln (2)}$

Step 1.2.6.5.3

Simplify the right side.

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Step 1.2.6.5.3.1

Move the negative in front of the fraction.

$x = - \frac{\ln (\frac{1}{5})}{\ln (2)}$

Step 1.3

x-intercept(s) in point form.

x-intercept(s): $(- \frac{\ln (\frac{1}{5})}{\ln (2)}, 0)$

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 $y$ .

$y = {(\frac{1}{2})}^{- (0)} - 5$

Step 2.2

Solve the equation.

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Step 2.2.1

Remove parentheses.

$y = {(\frac{1}{2})}^{- (0)} - 5$

Step 2.2.2

Simplify ${(\frac{1}{2})}^{- (0)} - 5$ .

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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 $0$ .

$y = {(\frac{1}{2})}^{0} - 5$

Step 2.2.2.1.2

Apply the product rule to $\frac{1}{2}$ .

$y = \frac{1^{0}}{2^{0}} - 5$

Step 2.2.2.1.3

Anything raised to $0$ is $1$ .

$y = \frac{1}{2^{0}} - 5$

Step 2.2.2.1.4

Anything raised to $0$ is $1$ .

$y = \frac{1}{1} - 5$

Step 2.2.2.1.5

Divide $1$ by $1$ .

$y = 1 - 5$

Step 2.2.2.2

Subtract $5$ from $1$ .

$y = - 4$

Step 2.3

y-intercept(s) in point form.

y-intercept(s): $(0, - 4)$

Step 3

List the intersections.

x-intercept(s): $(- \frac{\ln (\frac{1}{5})}{\ln (2)}, 0)$

y-intercept(s): $(0, - 4)$

Step 4