Precalculus Examples

$2^{x} \cdot 5^{x} = 100$

Step 1

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

$\ln (2^{x} \cdot 5^{x}) = \ln (100)$

Step 2

Expand the left side.

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Step 2.1

Rewrite $\ln (2^{x} \cdot 5^{x})$ as $\ln (2^{x}) + \ln (5^{x})$ .

$\ln (2^{x}) + \ln (5^{x}) = \ln (100)$

Step 2.2

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

$x \ln (2) + \ln (5^{x}) = \ln (100)$

Step 2.3

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

$x \ln (2) + x \ln (5) = \ln (100)$

Step 3

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

$x \ln (2) + x \ln (5) - \ln (100) = 0$

Step 4

Add $\ln (100)$ to both sides of the equation.

$x \ln (2) + x \ln (5) = \ln (100)$

Step 5

Factor $x$ out of $x \ln (2) + x \ln (5)$ .

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Step 5.1

Factor $x$ out of $x \ln (2)$ .

$x (\ln (2)) + x \ln (5) = \ln (100)$

Step 5.2

Factor $x$ out of $x \ln (5)$ .

$x (\ln (2)) + x (\ln (5)) = \ln (100)$

Step 5.3

Factor $x$ out of $x (\ln (2)) + x (\ln (5))$ .

$x (\ln (2) + \ln (5)) = \ln (100)$

Step 6

Divide each term in $x (\ln (2) + \ln (5)) = \ln (100)$ by $\ln (2) + \ln (5)$ and simplify.

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Step 6.1

Divide each term in $x (\ln (2) + \ln (5)) = \ln (100)$ by $\ln (2) + \ln (5)$ .

$\frac{x (\ln (2) + \ln (5))}{\ln (2) + \ln (5)} = \frac{\ln (100)}{\ln (2) + \ln (5)}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $\ln (2) + \ln (5)$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{x (\ln (2) + \ln (5))}{\ln (2) + \ln (5)} = \frac{\ln (100)}{\ln (2) + \ln (5)}$

Step 6.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (100)}{\ln (2) + \ln (5)}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (100)}{\ln (2) + \ln (5)}$

Decimal Form:

$x = 2$