Solve for x natural log of square root of x+6=2

$\ln (\sqrt{x + 6}) = 2$

Step 1

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

$e^{\ln (\sqrt{x + 6})} = e^{2}$

Step 2

Rewrite $\ln (\sqrt{x + 6}) = 2$ 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 $b^{y} = x$ .

$e^{2} = \sqrt{x + 6}$

Step 3

Solve for $x$ .

Rewrite the equation as $\sqrt{x + 6} = e^{2}$ .

$\sqrt{x + 6} = e^{2}$

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

${\sqrt{x + 6}}^{2} = {(e^{2})}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 6}$ as ${(x + 6)}^{\frac{1}{2}}$ .

${({(x + 6)}^{\frac{1}{2}})}^{2} = {(e^{2})}^{2}$

Simplify the left side.

Simplify ${({(x + 6)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x + 6)}^{\frac{1}{2}})}^{2}$ .

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

${(x + 6)}^{\frac{1}{2} \cdot 2} = {(e^{2})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x + 6)}^{\frac{1}{2} \cdot 2} = {(e^{2})}^{2}$

Rewrite the expression.

${(x + 6)}^{1} = {(e^{2})}^{2}$

Simplify.

$x + 6 = {(e^{2})}^{2}$

Simplify the right side.

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

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

$x + 6 = e^{2 \cdot 2}$

Multiply $2$ by $2$ .

$x + 6 = e^{4}$

Subtract $6$ from both sides of the equation.

$x = e^{4} - 6$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = e^{4} - 6$

Decimal Form:

$x = 48.59815003 \dots$

Solve for x natural log of square root of x+6=2

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