Precalculus Examples

$\ln (z^{3}) - 2 \ln (z) = 2$

Step 1

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

$\ln (z^{3}) - 2 \ln (z) = 2$

Step 2

Simplify the left side.

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Step 2.1

Simplify $\ln (z^{3}) - 2 \ln (z)$ .

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Step 2.1.1

Simplify $- 2 \ln (z)$ by moving $2$ inside the logarithm.

$\ln (z^{3}) - \ln (z^{2}) = 2$

Step 2.1.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{z^{3}}{z^{2}}) = 2$

Step 2.1.3

Cancel the common factor of $z^{3}$ and $z^{2}$ .

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Step 2.1.3.1

Factor $z^{2}$ out of $z^{3}$ .

$\ln (\frac{z^{2} z}{z^{2}}) = 2$

Step 2.1.3.2

Cancel the common factors.

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Step 2.1.3.2.1

Multiply by $1$ .

$\ln (\frac{z^{2} z}{z^{2} \cdot 1}) = 2$

Step 2.1.3.2.2

Cancel the common factor.

$\ln (\frac{z^{2} z}{z^{2} \cdot 1}) = 2$

Step 2.1.3.2.3

Rewrite the expression.

$\ln (\frac{z}{1}) = 2$

Step 2.1.3.2.4

Divide $z$ by $1$ .

$\ln (z) = 2$

Step 3

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

$e^{\ln (z)} = e^{2}$

Step 4

Rewrite $\ln (z) = 2$ 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 $b^{y} = x$ .

$e^{2} = z$

Step 5

Rewrite the equation as $z = e^{2}$ .

$z = e^{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$z = e^{2}$

Decimal Form:

$z = 7.38905609 \dots$