Solve for x natural log of 2+ natural log of x=1

$\ln (2) + \ln (x) = 1$

Step 1

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

$\ln (2 x) = 1$

Step 2

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

$e^{\ln (2 x)} = e^{1}$

Step 3

Rewrite $\ln (2 x) = 1$ 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 $b^{y} = x$ .

$e^{1} = 2 x$

Step 4

Solve for $x$ .

Rewrite the equation as $2 x = e^{1}$ .

$2 x = e$

Divide each term in $2 x = e^{1}$ by $2$ and simplify.

Divide each term in $2 x = e^{1}$ by $2$ .

$\frac{2 x}{2} = \frac{e^{1}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{e^{1}}{2}$

Divide $x$ by $1$ .

$x = \frac{e^{1}}{2}$

Simplify the right side.

Simplify.

$x = \frac{e}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{e}{2}$

Decimal Form:

$x = 1.35914091 \dots$

Solve for x natural log of 2+ natural log of x=1

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