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

Question

Lincoln Anderson · Accepted Answer

#x=e^(1/8)# 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)

Lincoln Anderson · Answer

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

Elias Ackerman · Answer

undefined To solve the equation 8ln(x) = 1:

1. Divide both sides by 8: ln(x) = 1/8.
2. Rewrite the equation in exponential form: e^(ln(x)) = e^(1/8).
3. Since e^(ln(x)) = x, the equation becomes x = e^(1/8).
4. Use a calculator to find the approximate value of e^(1/8).
5. The solution for x is approximately x ≈ 1.201.