Precalculus Examples

$2^{x} \cdot 5 = 21$

Step 1

Divide each term in $2^{x} \cdot 5 = 21$ by $5$ and simplify.

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Step 1.1

Divide each term in $2^{x} \cdot 5 = 21$ by $5$ .

$\frac{2^{x} \cdot 5}{5} = \frac{21}{5}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $5$ .

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Step 1.2.1.1

Cancel the common factor.

$\frac{2^{x} \cdot 5}{5} = \frac{21}{5}$

Step 1.2.1.2

Divide $2^{x}$ by $1$ .

$2^{x} = \frac{21}{5}$

Step 2

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

$\ln (2^{x}) = \ln (\frac{21}{5})$

Step 3

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

$x \ln (2) = \ln (\frac{21}{5})$

Step 4

Divide each term in $x \ln (2) = \ln (\frac{21}{5})$ by $\ln (2)$ and simplify.

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Step 4.1

Divide each term in $x \ln (2) = \ln (\frac{21}{5})$ by $\ln (2)$ .

$\frac{x \ln (2)}{\ln (2)} = \frac{\ln (\frac{21}{5})}{\ln (2)}$

Step 4.2

Simplify the left side.

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Step 4.2.1

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

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Step 4.2.1.1

Cancel the common factor.

$\frac{x \ln (2)}{\ln (2)} = \frac{\ln (\frac{21}{5})}{\ln (2)}$

Step 4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (\frac{21}{5})}{\ln (2)}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (\frac{21}{5})}{\ln (2)}$

Decimal Form:

$x = 2.07038932 \dots$