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}. #