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

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

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

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

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

$e^{1} = \sqrt{x + 2}$

Solve for $x$ .

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

$\sqrt{x + 2} = e$

To remove the radical on the left side of the equation, square both sides of the equation.

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

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 2}$ as ${(x + 2)}^{\frac{1}{2}}$ .

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

Simplify the left side.

Simplify ${({(x + 2)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x + 2)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Multiply the exponents in ${(e^{1})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x + 2 = e^{1 \cdot 2}$

Multiply $2$ by $1$ .

$x + 2 = e^{2}$

Subtract $2$ from both sides of the equation.

$x = e^{2} - 2$

The result can be shown in multiple forms.

Exact Form:

$x = e^{2} - 2$

Decimal Form:

$x = 5.38905609 \dots$

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

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