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

2 Answers
Mar 7, 2018

x=e^(1/8)

Explanation:

We know that,
color(red)((1)log_ax=n<=>x=a^n)
So,
8ln(x)=1rArrln(x)=1/8
rArrlog_ex=1/8<=>x=e^(1/8), Applying(1)

Mar 7, 2018

"The solution is:" \qquad \qquad \ x \ = \ e^{1/8} \ = \ root[8]{e}.

Explanation:

"We can work as follows:"

\quad \ 8 lnx \ = \ 1. \quad \ color{blue}{ "now isolate the log term" \ rarr }

\quad \ lnx \ = \ 1/8 \qquad \ \ color{blue}{ "now maybe emphasize the base of the log" \ rarr }

\quad \ log_{e} x \ = \ 1/8 \quad \ \color{blue}{ "now rewrite this as an exponential equation, " }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \color{blue}{ "using Fundamental Property of Logarithms:" }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \color{blue}{ log_{b} x = color{red}{p} \quad hArr \quad b^color{red}{p} = x. \qquad rarr }

\quad \ e^{1/8} \ = \ x \qquad \ \color{blue}{ "this is our solution !!" }

\quad \ x \ = \ e^{1/8} \qquad \ \color{blue}{ "write it the other way around; we are done." }

\quad \ x \ = \ root[8]{e} \qquad \color{blue}{ "or write it without negative or fractional" }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \color{blue}{ "exponents, if you like." }

"So, we have our solution:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x \ = \ e^{1/8} \ = \ root[8]{e}.