Precalculus Examples

$7 + 6 \ln (x) = 9$

Step 1

Move all terms not containing $\ln (x)$ to the right side of the equation.

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Step 1.1

Subtract $7$ from both sides of the equation.

$6 \ln (x) = 9 - 7$

Step 1.2

Subtract $7$ from $9$ .

$6 \ln (x) = 2$

Step 2

Divide each term in $6 \ln (x) = 2$ by $6$ and simplify.

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Step 2.1

Divide each term in $6 \ln (x) = 2$ by $6$ .

$\frac{6 \ln (x)}{6} = \frac{2}{6}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $6$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{6 \ln (x)}{6} = \frac{2}{6}$

Step 2.2.1.2

Divide $\ln (x)$ by $1$ .

$\ln (x) = \frac{2}{6}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Cancel the common factor of $2$ and $6$ .

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Step 2.3.1.1

Factor $2$ out of $2$ .

$\ln (x) = \frac{2 (1)}{6}$

Step 2.3.1.2

Cancel the common factors.

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Step 2.3.1.2.1

Factor $2$ out of $6$ .

$\ln (x) = \frac{2 \cdot 1}{2 \cdot 3}$

Step 2.3.1.2.2

Cancel the common factor.

$\ln (x) = \frac{2 \cdot 1}{2 \cdot 3}$

Step 2.3.1.2.3

Rewrite the expression.

$\ln (x) = \frac{1}{3}$

Step 3

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (x)} = e^{\frac{1}{3}}$

Step 4

Rewrite $\ln (x) = \frac{1}{3}$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{\frac{1}{3}} = x$

Step 5

Rewrite the equation as $x = e^{\frac{1}{3}}$ .

$x = e^{\frac{1}{3}}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = e^{\frac{1}{3}}$

Decimal Form:

$x = 1.39561242 \dots$