# How do you solve ln (sqrt(x+9)) = 1?

Mar 28, 2016

$x \approx - 1.61$

#### Explanation:

$1$. Use the natural logarithmic property, ${\ln}_{\textcolor{p u r p \le}{b}} \left({\textcolor{p u r p \le}{b}}^{\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{x}}\right) = \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{x}$, to rewrite the right side of the equation.

$\ln \left(\sqrt{x + 9}\right) = 1$

$\ln \left(\sqrt{x + 9}\right) = \ln \left({e}^{1}\right)$

$2$. Since the equation now follows a "$\ln = \ln$" situation, where the bases are the same on both sides, rewrite the equation without the "ln" portion.

$\sqrt{x + 9} = e$

$3$. Solve for $x$.

$x + 9 = {e}^{2}$

$x = {e}^{2} - 9$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x \approx - 1.61 \textcolor{w h i t e}{\frac{a}{a}} |}}}$