Precalculus Examples

$\log_{5} (5^{x + 1} - 20) = x$

Step 1

Set the argument in $\log_{5} (5^{x + 1} - 20)$ greater than $0$ to find where the expression is defined.

$5^{x + 1} - 20 > 0$

Step 2

Solve for $x$ .

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Step 2.1

Add $20$ to both sides of the inequality.

$5^{x + 1} > 20$

Step 2.2

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

$\ln (5^{x + 1}) > \ln (20)$

Step 2.3

Expand $\ln (5^{x + 1})$ by moving $x + 1$ outside the logarithm.

$(x + 1) \ln (5) > \ln (20)$

Step 2.4

Simplify the left side.

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Step 2.4.1

Simplify $(x + 1) \ln (5)$ .

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Step 2.4.1.1

Apply the distributive property.

$x \ln (5) + 1 \ln (5) > \ln (20)$

Step 2.4.1.2

Multiply $\ln (5)$ by $1$ .

$x \ln (5) + \ln (5) > \ln (20)$

Step 2.5

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

$x \ln (5) + \ln (5) - \ln (20) = 0$

Step 2.6

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$x \ln (5) + \ln (\frac{5}{20}) = 0$

Step 2.7

Cancel the common factor of $5$ and $20$ .

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Step 2.7.1

Factor $5$ out of $5$ .

$x \ln (5) + \ln (\frac{5 (1)}{20}) = 0$

Step 2.7.2

Cancel the common factors.

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Step 2.7.2.1

Factor $5$ out of $20$ .

$x \ln (5) + \ln (\frac{5 \cdot 1}{5 \cdot 4}) = 0$

Step 2.7.2.2

Cancel the common factor.

$x \ln (5) + \ln (\frac{5 \cdot 1}{5 \cdot 4}) = 0$

Step 2.7.2.3

Rewrite the expression.

$x \ln (5) + \ln (\frac{1}{4}) = 0$

Step 2.8

Subtract $\ln (\frac{1}{4})$ from both sides of the equation.

$x \ln (5) = - \ln (\frac{1}{4})$

Step 2.9

Divide each term in $x \ln (5) = - \ln (\frac{1}{4})$ by $\ln (5)$ and simplify.

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Step 2.9.1

Divide each term in $x \ln (5) = - \ln (\frac{1}{4})$ by $\ln (5)$ .

$\frac{x \ln (5)}{\ln (5)} = \frac{- \ln (\frac{1}{4})}{\ln (5)}$

Step 2.9.2

Simplify the left side.

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Step 2.9.2.1

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

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Step 2.9.2.1.1

Cancel the common factor.

$\frac{x \ln (5)}{\ln (5)} = \frac{- \ln (\frac{1}{4})}{\ln (5)}$

Step 2.9.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \ln (\frac{1}{4})}{\ln (5)}$

Step 2.9.3

Simplify the right side.

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Step 2.9.3.1

Move the negative in front of the fraction.

$x = - \frac{\ln (\frac{1}{4})}{\ln (5)}$

Step 2.10

The solution consists of all of the true intervals.

$x > - \frac{\ln (\frac{1}{4})}{\ln (5)}$

Step 3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \frac{\ln (\frac{1}{4})}{\ln (5)}, \infty)$

Set-Builder Notation:

${x | x > - \frac{\ln (\frac{1}{4})}{\ln (5)}}$

Step 4