Solve for k natural log of 2+ natural log of k-4=1/3

$\ln (2) + \ln (k - 4) = \frac{1}{3}$

Step 1

Simplify the left side.

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\ln (2 (k - 4)) = \frac{1}{3}$

Apply the distributive property.

$\ln (2 k + 2 \cdot - 4) = \frac{1}{3}$

Multiply $2$ by $- 4$ .

$\ln (2 k - 8) = \frac{1}{3}$

Step 2

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

$e^{\ln (2 k - 8)} = e^{\frac{1}{3}}$

Step 3

Rewrite $\ln (2 k - 8) = \frac{1}{3}$ 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 $b^{y} = x$ .

$e^{\frac{1}{3}} = 2 k - 8$

Step 4

Solve for $k$ .

Rewrite the equation as $2 k - 8 = e^{\frac{1}{3}}$ .

$2 k - 8 = e^{\frac{1}{3}}$

Add $8$ to both sides of the equation.

$2 k = e^{\frac{1}{3}} + 8$

Divide each term in $2 k = e^{\frac{1}{3}} + 8$ by $2$ and simplify.

Divide each term in $2 k = e^{\frac{1}{3}} + 8$ by $2$ .

$\frac{2 k}{2} = \frac{e^{\frac{1}{3}}}{2} + \frac{8}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 k}{2} = \frac{e^{\frac{1}{3}}}{2} + \frac{8}{2}$

Divide $k$ by $1$ .

$k = \frac{e^{\frac{1}{3}}}{2} + \frac{8}{2}$

Simplify the right side.

Divide $8$ by $2$ .

$k = \frac{e^{\frac{1}{3}}}{2} + 4$

Step 5

The result can be shown in multiple forms.

Exact Form:

$k = \frac{e^{\frac{1}{3}}}{2} + 4$

Decimal Form:

$k = 4.69780621 \dots$

Solve for k natural log of 2+ natural log of k-4=1/3

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