Precalculus Examples

$5^{2 x} + 5^{x + 1} - 14 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Rewrite $5^{x + 1}$ as $5^{x} \cdot 5^{1}$ .

$5^{2 x} + 5^{x} \cdot 5 - 14 = 0$

Step 1.2

Rewrite $5^{2 x}$ as ${(5^{x})}^{2}$ .

${(5^{x})}^{2} + 5^{x} \cdot 5 - 14 = 0$

Step 1.3

Let $u = 5^{x}$ . Substitute $u$ for all occurrences of $5^{x}$ .

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Step 1.3.1

Evaluate the exponent.

$u^{2} + u \cdot 5 - 14 = 0$

Step 1.3.2

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

$u^{2} + 5 u - 14 = 0$

Step 1.4

Factor $u^{2} + 5 u - 14$ using the AC method.

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Step 1.4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 14$ and whose sum is $5$ .

$- 2, 7$

Step 1.4.2

Write the factored form using these integers.

$(u - 2) (u + 7) = 0$

Step 1.5

Replace all occurrences of $u$ with $5^{x}$ .

$(5^{x} - 2) (5^{x} + 7) = 0$

Step 2

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

$5^{x} - 2 = 0$

$5^{x} + 7 = 0$

Step 3

Set $5^{x} - 2$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $5^{x} - 2$ equal to $0$ .

$5^{x} - 2 = 0$

Step 3.2

Solve $5^{x} - 2 = 0$ for $x$ .

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Step 3.2.1

Add $2$ to both sides of the equation.

$5^{x} = 2$

Step 3.2.2

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

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

Step 3.2.3

Expand $\ln (5^{x})$ by moving $x$ outside the logarithm.

$x \ln (5) = \ln (2)$

Step 3.2.4

Divide each term in $x \ln (5) = \ln (2)$ by $\ln (5)$ and simplify.

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Step 3.2.4.1

Divide each term in $x \ln (5) = \ln (2)$ by $\ln (5)$ .

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

Step 3.2.4.2

Simplify the left side.

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Step 3.2.4.2.1

Cancel the common factor of $\ln (5)$ .

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Step 3.2.4.2.1.1

Cancel the common factor.

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

Step 3.2.4.2.1.2

Divide $x$ by $1$ .

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

Step 4

Set $5^{x} + 7$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $5^{x} + 7$ equal to $0$ .

$5^{x} + 7 = 0$

Step 4.2

Solve $5^{x} + 7 = 0$ for $x$ .

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Step 4.2.1

Subtract $7$ from both sides of the equation.

$5^{x} = - 7$

Step 4.2.2

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

$\ln (5^{x}) = \ln (- 7)$

Step 4.2.3

The equation cannot be solved because $\ln (- 7)$ is undefined.

Undefined

Step 4.2.4

There is no solution for $5^{x} = - 7$

No solution

Step 5

The final solution is all the values that make $(5^{x} - 2) (5^{x} + 7) = 0$ true.

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

Step 6

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.43067655 \dots$