# How do you solve 8ln(x) = 1?

Mar 7, 2018

$x = {e}^{\frac{1}{8}}$

#### Explanation:

We know that,
$\textcolor{red}{\left(1\right) {\log}_{a} x = n \iff x = {a}^{n}}$
So,
$8 \ln \left(x\right) = 1 \Rightarrow \ln \left(x\right) = \frac{1}{8}$
$\Rightarrow {\log}_{e} x = \frac{1}{8} \iff x = {e}^{\frac{1}{8}}$, Applying(1)

Mar 7, 2018

$\text{The solution is:} \setminus q \quad \setminus q \quad \setminus x \setminus = \setminus {e}^{\frac{1}{8}} \setminus = \setminus \sqrt[8]{e} .$

#### Explanation:

$\text{We can work as follows:}$

$\setminus \quad \setminus 8 \ln x \setminus = \setminus 1. \setminus \quad \setminus \textcolor{b l u e}{\text{now isolate the log term} \setminus \rightarrow}$

$\setminus \quad \setminus \ln x \setminus = \setminus \frac{1}{8} \setminus q \quad \setminus \setminus \textcolor{b l u e}{\text{now maybe emphasize the base of the log} \setminus \rightarrow}$

$\setminus \quad \setminus {\log}_{e} x \setminus = \setminus \frac{1}{8} \setminus \quad \setminus \setminus \textcolor{b l u e}{\text{now rewrite this as an exponential equation, }}$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \textcolor{b l u e}{\text{using Fundamental Property of Logarithms:}}$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \textcolor{b l u e}{{\log}_{b} x = \textcolor{red}{p} \setminus \quad \Leftrightarrow \setminus \quad {b}^{\textcolor{red}{p}} = x . \setminus q \quad \rightarrow}$

$\setminus \quad \setminus {e}^{\frac{1}{8}} \setminus = \setminus x \setminus q \quad \setminus \setminus \textcolor{b l u e}{\text{this is our solution !!}}$

$\setminus \quad \setminus x \setminus = \setminus {e}^{\frac{1}{8}} \setminus q \quad \setminus \setminus \textcolor{b l u e}{\text{write it the other way around; we are done.}}$

$\setminus \quad \setminus x \setminus = \setminus \sqrt[8]{e} \setminus q \quad \setminus \textcolor{b l u e}{\text{or write it without negative or fractional}}$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus \textcolor{b l u e}{\text{exponents, if you like.}}$

$\text{So, we have our solution:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad x \setminus = \setminus {e}^{\frac{1}{8}} \setminus = \setminus \sqrt[8]{e} .$