Question 251116
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(\sqrt[3]{\left(9x\,-\,8\right)^2}\right)\ =\ \frac{4}{3}]


I presume base 10 because you didn't specify a base.


Use rational exponents, which is to say use the rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^{\frac{1}{n}} = \sqrt[n]{a}]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(\left(9x\,-\,8\right)^{\frac{2}{3}}\right)\ =\ \frac{4}{3}]


Now use the rule of logarithms that says:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n)\ =\ n\log_b(x)]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{2}{3}\right)\log\left(9x\,-\,8\right)\ =\ \frac{4}{3}]


Multiply both sides of the equation by *[tex \LARGE \frac{3}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(9x\,-\,8\right)\ =\ 2]


Use the definition of logarithms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \log_b(x) \ \ \Rightarrow\ \ b^y\ =\ x]


To write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10^2\ =\ 9x\ -\ 8]


Then solve the simple linear equation for *[tex \Large x]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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But aren't there two solutions John?