Solve for x e^( natural log of x+1)=2

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

Step 1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

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

Step 2

Expand the left side.

Expand $\ln (e^{\ln (x) + 1})$ by moving $\ln (x) + 1$ outside the logarithm.

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

The natural logarithm of $e$ is $1$ .

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

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

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

Step 3

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

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

Step 4

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

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

Step 5

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

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

Step 6

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

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

Step 7

Solve for $x$ .

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

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

Multiply both sides of the equation by $2$ .

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

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$x = 2 e^{- 1}$

Simplify the right side.

Simplify $2 e^{- 1}$ .

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

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

Combine $2$ and $\frac{1}{e}$ .

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

Step 8

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.73575888 \dots$

Solve for x e^( natural log of x+1)=2

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