Finite Math Examples

$15 = 5 {(1.07)}^{x \cdot 4}$

Step 1

Rewrite the equation as $5 \cdot {1.07}^{x \cdot 4} = 15$ .

$5 \cdot {1.07}^{x \cdot 4} = 15$

Step 2

Divide each term in $5 \cdot {1.07}^{x \cdot 4} = 15$ by $5$ and simplify.

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Step 2.1

Divide each term in $5 \cdot {1.07}^{x \cdot 4} = 15$ by $5$ .

$\frac{5 \cdot {1.07}^{x \cdot 4}}{5} = \frac{15}{5}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $5$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{5 \cdot {1.07}^{x \cdot 4}}{5} = \frac{15}{5}$

Step 2.2.1.2

Divide ${1.07}^{x \cdot 4}$ by $1$ .

${1.07}^{x \cdot 4} = \frac{15}{5}$

Step 2.2.2

Move $4$ to the left of $x$ .

${1.07}^{4 x} = \frac{15}{5}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Divide $15$ by $5$ .

${1.07}^{4 x} = 3$

Step 3

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

$\ln ({1.07}^{4 x}) = \ln (3)$

Step 4

Expand $\ln ({1.07}^{4 x})$ by moving $4 x$ outside the logarithm.

$4 x \ln (1.07) = \ln (3)$

Step 5

Divide each term in $4 x \ln (1.07) = \ln (3)$ by $4 \ln (1.07)$ and simplify.

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Step 5.1

Divide each term in $4 x \ln (1.07) = \ln (3)$ by $4 \ln (1.07)$ .

$\frac{4 x \ln (1.07)}{4 \ln (1.07)} = \frac{\ln (3)}{4 \ln (1.07)}$

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $4$ .

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Step 5.2.1.1

Cancel the common factor.

$\frac{4 x \ln (1.07)}{4 \ln (1.07)} = \frac{\ln (3)}{4 \ln (1.07)}$

Step 5.2.1.2

Rewrite the expression.

$\frac{x \ln (1.07)}{\ln (1.07)} = \frac{\ln (3)}{4 \ln (1.07)}$

Step 5.2.2

Cancel the common factor of $\ln (1.07)$ .

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Step 5.2.2.1

Cancel the common factor.

$\frac{x \ln (1.07)}{\ln (1.07)} = \frac{\ln (3)}{4 \ln (1.07)}$

Step 5.2.2.2

Divide $x$ by $1$ .

$x = \frac{\ln (3)}{4 \ln (1.07)}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (3)}{4 \ln (1.07)}$

Decimal Form:

$x = 4.05939341 \dots$