Question 422460
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First use the fact that 8 is an integer power of 2, that is:  *[tex \Large 8\ =\ 2^3], and the fact that *[tex \Large x^{a^b}\ =\ x^{ab}] to rewrite your equation as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^{9x\ -\ 18}\ =\ 2^{4x\ -\ 5}]


Then take the natural logarithm of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(2^{9x\ -\ 18}\right)\ =\ \ln\left(2^{4x\ -\ 5}\right)]


Then use:


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (9x\ -\ 18)\ln\left(2\right)\ =\ (4x\ -\ 5)\ln\left(2\right)]


Divide by *[tex \Large \ln(2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (9x\ -\ 18)\ =\ (4x\ -\ 5)]


and finally solve for *[tex \Large x]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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