Solve for x natural log of 5=1+ natural log of 2x

$\ln (5) = 1 + \ln (2 x)$

Step 1

Rewrite the equation as $1 + \ln (2 x) = \ln (5)$ .

$1 + \ln (2 x) = \ln (5)$

Step 2

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

$\ln (2 x) - \ln (5) = - 1$

Step 3

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

$\ln (\frac{2 x}{5}) = - 1$

Step 4

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

$e^{\ln (\frac{2 x}{5})} = e^{- 1}$

Step 5

Rewrite $\ln (\frac{2 x}{5}) = - 1$ 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 $b^{y} = x$ .

$e^{- 1} = \frac{2 x}{5}$

Step 6

Solve for $x$ .

Rewrite the equation as $\frac{2 x}{5} = e^{- 1}$ .

$\frac{2 x}{5} = e^{- 1}$

Multiply both sides of the equation by $\frac{5}{2}$ .

$\frac{5}{2} \cdot \frac{2 x}{5} = \frac{5}{2} e^{- 1}$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{5}{2} \cdot \frac{2 x}{5}$ .

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5}{2} \cdot \frac{2 x}{5} = \frac{5}{2} e^{- 1}$

Rewrite the expression.

$\frac{1}{2} (2 x) = \frac{5}{2} e^{- 1}$

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

$\frac{1}{2} (2 (x)) = \frac{5}{2} e^{- 1}$

Cancel the common factor.

$\frac{1}{2} (2 x) = \frac{5}{2} e^{- 1}$

Rewrite the expression.

$x = \frac{5}{2} e^{- 1}$

Simplify the right side.

Simplify $\frac{5}{2} e^{- 1}$ .

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$x = \frac{5}{2} \cdot \frac{1}{e}$

Multiply $\frac{5}{2}$ by $\frac{1}{e}$ .

$x = \frac{5}{2 e}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \frac{5}{2 e}$

Decimal Form:

$x = 0.91969860 \dots$

Solve for x natural log of 5=1+ natural log of 2x

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