Find the Exponential Function (-3,125)

$(- 3, 125)$

Step 1

To find an exponential function, $f (x) = a^{x}$ , containing the point, set $f (x)$ in the function to the $y$ value $125$ of the point, and set $x$ to the $x$ value $- 3$ of the point.

$125 = a^{- 3}$

Step 2

Solve the equation for $a$ .

Rewrite the equation as $a^{- 3} = 125$ .

$a^{- 3} = 125$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{a^{3}} = 125$

Find the LCD of the terms in the equation.

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

$a^{3}, 1$

The LCM of one and any expression is the expression.

$a^{3}$

Multiply each term in $\frac{1}{a^{3}} = 125$ by $a^{3}$ to eliminate the fractions.

Multiply each term in $\frac{1}{a^{3}} = 125$ by $a^{3}$ .

$\frac{1}{a^{3}} a^{3} = 125 a^{3}$

Simplify the left side.

Cancel the common factor of $a^{3}$ .

Cancel the common factor.

$\frac{1}{a^{3}} a^{3} = 125 a^{3}$

Rewrite the expression.

$1 = 125 a^{3}$

Solve the equation.

Rewrite the equation as $125 a^{3} = 1$ .

$125 a^{3} = 1$

Subtract $1$ from both sides of the equation.

$125 a^{3} - 1 = 0$

Factor the left side of the equation.

Rewrite $125 a^{3}$ as ${(5 a)}^{3}$ .

${(5 a)}^{3} - 1 = 0$

Rewrite $1$ as $1^{3}$ .

${(5 a)}^{3} - 1^{3} = 0$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = 5 a$ and $b = 1$ .

$(5 a - 1) ({(5 a)}^{2} + 5 a \cdot 1 + 1^{2}) = 0$

Simplify.

Apply the product rule to $5 a$ .

$(5 a - 1) (5^{2} a^{2} + 5 a \cdot 1 + 1^{2}) = 0$

Raise $5$ to the power of $2$ .

$(5 a - 1) (25 a^{2} + 5 a \cdot 1 + 1^{2}) = 0$

Multiply $5$ by $1$ .

$(5 a - 1) (25 a^{2} + 5 a + 1^{2}) = 0$

One to any power is one.

$(5 a - 1) (25 a^{2} + 5 a + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$5 a - 1 = 0$

$25 a^{2} + 5 a + 1 = 0$

Set $5 a - 1$ equal to $0$ and solve for $a$ .

Set $5 a - 1$ equal to $0$ .

$5 a - 1 = 0$

Solve $5 a - 1 = 0$ for $a$ .

Add $1$ to both sides of the equation.

$5 a = 1$

Divide each term in $5 a = 1$ by $5$ and simplify.

Divide each term in $5 a = 1$ by $5$ .

$\frac{5 a}{5} = \frac{1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 a}{5} = \frac{1}{5}$

Divide $a$ by $1$ .

$a = \frac{1}{5}$

Set $25 a^{2} + 5 a + 1$ equal to $0$ and solve for $a$ .

Set $25 a^{2} + 5 a + 1$ equal to $0$ .

$25 a^{2} + 5 a + 1 = 0$

Solve $25 a^{2} + 5 a + 1 = 0$ for $a$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 25$ , $b = 5$ , and $c = 1$ into the quadratic formula and solve for $a$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (25 \cdot 1)}}{2 \cdot 25}$

Simplify.

Simplify the numerator.

Raise $5$ to the power of $2$ .

$a = \frac{- 5 \pm \sqrt{25 - 4 \cdot 25 \cdot 1}}{2 \cdot 25}$

Multiply $- 4 \cdot 25 \cdot 1$ .

Multiply $- 4$ by $25$ .

$a = \frac{- 5 \pm \sqrt{25 - 100 \cdot 1}}{2 \cdot 25}$

Multiply $- 100$ by $1$ .

$a = \frac{- 5 \pm \sqrt{25 - 100}}{2 \cdot 25}$

Subtract $100$ from $25$ .

$a = \frac{- 5 \pm \sqrt{- 75}}{2 \cdot 25}$

Rewrite $- 75$ as $- 1 (75)$ .

$a = \frac{- 5 \pm \sqrt{- 1 \cdot 75}}{2 \cdot 25}$

Rewrite $\sqrt{- 1 (75)}$ as $\sqrt{- 1} \cdot \sqrt{75}$ .

$a = \frac{- 5 \pm \sqrt{- 1} \cdot \sqrt{75}}{2 \cdot 25}$

Rewrite $\sqrt{- 1}$ as $i$ .

$a = \frac{- 5 \pm i \cdot \sqrt{75}}{2 \cdot 25}$

Rewrite $75$ as $5^{2} \cdot 3$ .

Factor $25$ out of $75$ .

$a = \frac{- 5 \pm i \cdot \sqrt{25 (3)}}{2 \cdot 25}$

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

$a = \frac{- 5 \pm i \cdot \sqrt{5^{2} \cdot 3}}{2 \cdot 25}$

Pull terms out from under the radical.

$a = \frac{- 5 \pm i \cdot (5 \sqrt{3})}{2 \cdot 25}$

Move $5$ to the left of $i$ .

$a = \frac{- 5 \pm 5 i \sqrt{3}}{2 \cdot 25}$

Multiply $2$ by $25$ .

$a = \frac{- 5 \pm 5 i \sqrt{3}}{50}$

Simplify $\frac{- 5 \pm 5 i \sqrt{3}}{50}$ .

$a = \frac{- 1 \pm i \sqrt{3}}{10}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $5$ to the power of $2$ .

$a = \frac{- 5 \pm \sqrt{25 - 4 \cdot 25 \cdot 1}}{2 \cdot 25}$

Multiply $- 4 \cdot 25 \cdot 1$ .

Multiply $- 4$ by $25$ .

$a = \frac{- 5 \pm \sqrt{25 - 100 \cdot 1}}{2 \cdot 25}$

Multiply $- 100$ by $1$ .

$a = \frac{- 5 \pm \sqrt{25 - 100}}{2 \cdot 25}$

Subtract $100$ from $25$ .

$a = \frac{- 5 \pm \sqrt{- 75}}{2 \cdot 25}$

Rewrite $- 75$ as $- 1 (75)$ .

$a = \frac{- 5 \pm \sqrt{- 1 \cdot 75}}{2 \cdot 25}$

Rewrite $\sqrt{- 1 (75)}$ as $\sqrt{- 1} \cdot \sqrt{75}$ .

$a = \frac{- 5 \pm \sqrt{- 1} \cdot \sqrt{75}}{2 \cdot 25}$

Rewrite $\sqrt{- 1}$ as $i$ .

$a = \frac{- 5 \pm i \cdot \sqrt{75}}{2 \cdot 25}$

Rewrite $75$ as $5^{2} \cdot 3$ .

Factor $25$ out of $75$ .

$a = \frac{- 5 \pm i \cdot \sqrt{25 (3)}}{2 \cdot 25}$

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

$a = \frac{- 5 \pm i \cdot \sqrt{5^{2} \cdot 3}}{2 \cdot 25}$

Pull terms out from under the radical.

$a = \frac{- 5 \pm i \cdot (5 \sqrt{3})}{2 \cdot 25}$

Move $5$ to the left of $i$ .

$a = \frac{- 5 \pm 5 i \sqrt{3}}{2 \cdot 25}$

Multiply $2$ by $25$ .

$a = \frac{- 5 \pm 5 i \sqrt{3}}{50}$

Simplify $\frac{- 5 \pm 5 i \sqrt{3}}{50}$ .

$a = \frac{- 1 \pm i \sqrt{3}}{10}$

Change the $\pm$ to $+$ .

$a = \frac{- 1 + i \sqrt{3}}{10}$

Rewrite $- 1$ as $- 1 (1)$ .

$a = \frac{- 1 \cdot 1 + i \sqrt{3}}{10}$

Factor $- 1$ out of $i \sqrt{3}$ .

$a = \frac{- 1 \cdot 1 - (- i \sqrt{3})}{10}$

Factor $- 1$ out of $- 1 (1) - (- i \sqrt{3})$ .

$a = \frac{- 1 (1 - i \sqrt{3})}{10}$

Move the negative in front of the fraction.

$a = - \frac{1 - i \sqrt{3}}{10}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $5$ to the power of $2$ .

$a = \frac{- 5 \pm \sqrt{25 - 4 \cdot 25 \cdot 1}}{2 \cdot 25}$

Multiply $- 4 \cdot 25 \cdot 1$ .

Multiply $- 4$ by $25$ .

$a = \frac{- 5 \pm \sqrt{25 - 100 \cdot 1}}{2 \cdot 25}$

Multiply $- 100$ by $1$ .

$a = \frac{- 5 \pm \sqrt{25 - 100}}{2 \cdot 25}$

Subtract $100$ from $25$ .

$a = \frac{- 5 \pm \sqrt{- 75}}{2 \cdot 25}$

Rewrite $- 75$ as $- 1 (75)$ .

$a = \frac{- 5 \pm \sqrt{- 1 \cdot 75}}{2 \cdot 25}$

Rewrite $\sqrt{- 1 (75)}$ as $\sqrt{- 1} \cdot \sqrt{75}$ .

$a = \frac{- 5 \pm \sqrt{- 1} \cdot \sqrt{75}}{2 \cdot 25}$

Rewrite $\sqrt{- 1}$ as $i$ .

$a = \frac{- 5 \pm i \cdot \sqrt{75}}{2 \cdot 25}$

Rewrite $75$ as $5^{2} \cdot 3$ .

Factor $25$ out of $75$ .

$a = \frac{- 5 \pm i \cdot \sqrt{25 (3)}}{2 \cdot 25}$

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

$a = \frac{- 5 \pm i \cdot \sqrt{5^{2} \cdot 3}}{2 \cdot 25}$

Pull terms out from under the radical.

$a = \frac{- 5 \pm i \cdot (5 \sqrt{3})}{2 \cdot 25}$

Move $5$ to the left of $i$ .

$a = \frac{- 5 \pm 5 i \sqrt{3}}{2 \cdot 25}$

Multiply $2$ by $25$ .

$a = \frac{- 5 \pm 5 i \sqrt{3}}{50}$

Simplify $\frac{- 5 \pm 5 i \sqrt{3}}{50}$ .

$a = \frac{- 1 \pm i \sqrt{3}}{10}$

Change the $\pm$ to $-$ .

$a = \frac{- 1 - i \sqrt{3}}{10}$

Rewrite $- 1$ as $- 1 (1)$ .

$a = \frac{- 1 \cdot 1 - i \sqrt{3}}{10}$

Factor $- 1$ out of $- i \sqrt{3}$ .

$a = \frac{- 1 \cdot 1 - (i \sqrt{3})}{10}$

Factor $- 1$ out of $- 1 (1) - (i \sqrt{3})$ .

$a = \frac{- 1 (1 + i \sqrt{3})}{10}$

Move the negative in front of the fraction.

$a = - \frac{1 + i \sqrt{3}}{10}$

The final answer is the combination of both solutions.

$a = - \frac{1 - i \sqrt{3}}{10}, - \frac{1 + i \sqrt{3}}{10}$

The final solution is all the values that make $(5 a - 1) (25 a^{2} + 5 a + 1) = 0$ true.

$a = \frac{1}{5}, - \frac{1 - i \sqrt{3}}{10}, - \frac{1 + i \sqrt{3}}{10}$

Remove all values that contain imaginary components.

There are no imaginary components. Add $\frac{1}{5}$ to the final answer.

$\frac{1}{5}$ is a real number

The letter $i$ represents an imaginary component, and is not a real number. Do not add $- \frac{1 - i \sqrt{3}}{10}$ to the final answer.

$- \frac{1 - i \sqrt{3}}{10}$ is not a real number

The letter $i$ represents an imaginary component, and is not a real number. Do not add $- \frac{1 + i \sqrt{3}}{10}$ to the final answer.

$- \frac{1 + i \sqrt{3}}{10}$ is not a real number

The final answer is the list of values not containing imaginary components.

$a = \frac{1}{5}$

Step 3

Substitute each value for $a$ back into the function $f (x) = a^{x}$ to find each possible exponential function.

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

Find the Exponential Function (-3,125)

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