Algebra Examples

$(0, - 5)$ and $(4, - 80)$

Step 1

Substitute the values from the given points into the formula for an exponential function, $y = a b^{x}$ .

$- 5 = a b^{0}$

$- 80 = a b^{4}$

Step 2

Simplify the right side.

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Step 2.1

Simplify $a b^{0}$ .

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Step 2.1.1

Anything raised to $0$ is $1$ .

$- 5 = a \cdot 1$

$- 80 = a b^{4}$

Step 2.1.2

Multiply $a$ by $1$ .

$- 5 = a$

$- 80 = a b^{4}$

$- 5 = a$

$- 80 = a b^{4}$

$- 5 = a$

$- 80 = a b^{4}$

Step 3

Solve the two equations for $a$ and $b$ .

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Step 3.1

Rewrite the equation as $a = - 5$ .

$a = - 5$

$- 80 = a b^{4}$

Step 3.2

Replace all occurrences of $a$ with $- 5$ in each equation.

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Step 3.2.1

Replace all occurrences of $a$ in $- 80 = a b^{4}$ with $- 5$ .

$- 80 = (- 5) \cdot b^{4}$

$a = - 5$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Multiply $- 5$ by $b^{4}$ .

$- 80 = - 5 b^{4}$

$a = - 5$

$- 80 = - 5 b^{4}$

$a = - 5$

$- 80 = - 5 b^{4}$

$a = - 5$

Step 3.3

Solve for $b$ in $- 80 = - 5 b^{4}$ .

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Step 3.3.1

Rewrite the equation as $- 5 b^{4} = - 80$ .

$- 5 b^{4} = - 80$

$a = - 5$

Step 3.3.2

Divide each term in $- 5 b^{4} = - 80$ by $- 5$ and simplify.

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Step 3.3.2.1

Divide each term in $- 5 b^{4} = - 80$ by $- 5$ .

$\frac{- 5 b^{4}}{- 5} = \frac{- 80}{- 5}$

$a = - 5$

Step 3.3.2.2

Simplify the left side.

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Step 3.3.2.2.1

Cancel the common factor of $- 5$ .

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Step 3.3.2.2.1.1

Cancel the common factor.

$\frac{- 5 b^{4}}{- 5} = \frac{- 80}{- 5}$

$a = - 5$

Step 3.3.2.2.1.2

Divide $b^{4}$ by $1$ .

$b^{4} = \frac{- 80}{- 5}$

$a = - 5$

$b^{4} = \frac{- 80}{- 5}$

$a = - 5$

$b^{4} = \frac{- 80}{- 5}$

$a = - 5$

Step 3.3.2.3

Simplify the right side.

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Step 3.3.2.3.1

Divide $- 80$ by $- 5$ .

$b^{4} = 16$

$a = - 5$

$b^{4} = 16$

$a = - 5$

$b^{4} = 16$

$a = - 5$

Step 3.3.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$b = \pm \sqrt[4]{16}$

$a = - 5$

Step 3.3.4

Simplify $\pm \sqrt[4]{16}$ .

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Step 3.3.4.1

Rewrite $16$ as $2^{4}$ .

$b = \pm \sqrt[4]{2^{4}}$

$a = - 5$

Step 3.3.4.2

Pull terms out from under the radical, assuming positive real numbers.

$b = \pm 2$

$a = - 5$

$b = \pm 2$

$a = - 5$

Step 3.3.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.3.5.1

First, use the positive value of the $\pm$ to find the first solution.

$b = 2$

$a = - 5$

Step 3.3.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$b = - 2$

$a = - 5$

Step 3.3.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$b = 2, - 2$

$a = - 5$

$b = 2, - 2$

$a = - 5$

$b = 2, - 2$

$a = - 5$

Step 3.4

Solve the system of equations.

$b = 2, a = - 5$

Step 3.5

Solve the system of equations.

$b = - 2, a = - 5$

Step 3.6

List all of the solutions.

$b = 2, a = - 5$

$b = - 2, a = - 5$

$b = 2, a = - 5$

$b = - 2, a = - 5$

Step 4

Substitute the values found for $a$ and $b$ into $y = a b^{x}$ to find the exponential, ignoring negative numbers for $b$ .

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

Step 5