Question 251116
Assuming you are using a log of base 10.

{{{log (((9x-8)^2)^(1/3)) = 4/3}}} "^(1/3)" is another way of saying cube root.

{{{(1/3) * log((9x-8)^2) = 4/3}}} using log rule 3
{{{ log ((9x-8)^2) = 4}}} divide both sides by 1/3, (4/3)/(1/3) = (4/3)*(3/1)
{{{ (9x-8)^2 = 10000}}} inverse of log base 10 is 10^x,10^4 = 10000
take the square root of both sides, remember the roots will be negative and positive
{{{ (9x-8) = 100}}} OR {{{(9x-8) = -100}}}
Rearrange to find x
{{{ x = 12}}} OR {{{x= -(92/9)}}}

Does that make sense?