Precalculus Examples

$7 = 1 + 1.3 \ln (x + 1)$

Step 1

Rewrite the equation as $1 + 1.3 \ln (x + 1) = 7$ .

$1 + 1.3 \ln (x + 1) = 7$

Step 2

Move all terms not containing $\ln (x + 1)$ to the right side of the equation.

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Step 2.1

Subtract $1$ from both sides of the equation.

$1.3 \ln (x + 1) = 7 - 1$

Step 2.2

Subtract $1$ from $7$ .

$1.3 \ln (x + 1) = 6$

Step 3

Divide each term in $1.3 \ln (x + 1) = 6$ by $1.3$ and simplify.

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Step 3.1

Divide each term in $1.3 \ln (x + 1) = 6$ by $1.3$ .

$\frac{1.3 \ln (x + 1)}{1.3} = \frac{6}{1.3}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $1.3$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{1.3 \ln (x + 1)}{1.3} = \frac{6}{1.3}$

Step 3.2.1.2

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

$\ln (x + 1) = \frac{6}{1.3}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Divide $6$ by $1.3$ .

$\ln (x + 1) = 4.\overline{615384}$

Step 4

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

$e^{\ln (x + 1)} = e^{4.\overline{615384}}$

Step 5

Rewrite $\ln (x + 1) = 4.\overline{615384}$ 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 $b^{y} = x$ .

$e^{4.\overline{615384}} = x + 1$

Step 6

Solve for $x$ .

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Step 6.1

Rewrite the equation as $x + 1 = e^{4.\overline{615384}}$ .

$x + 1 = e^{4.\overline{615384}}$

Step 6.2

Subtract $1$ from both sides of the equation.

$x = e^{4.\overline{615384}} - 1$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = e^{4.\overline{615384}} - 1$

Decimal Form:

$x = 100.02667747 \dots$