Question 753700
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You are stuck because you failed to take the log of the RHS when you took the log of the LHS.  Furthermore, you didn't take the base 3 log of the LHS correctly.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3^{\large{2x-4}\LARGE}\ +\ 5\ =\ 70]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3^{\large{2x-4}\LARGE}\ =\ 65]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\left(3^{\large{2x-4}\LARGE}\right)\ =\ \log_3\left(65\right)]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2x\ -\ 4)\log_3\left(3\right)\ =\ \log_3\left(65\right)]


Now use *[tex \LARGE \log_b(b)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ -\ 4\ =\ \log_3\left(65\right)]


From here, just solve for *[tex \LARGE x] remembering that the log of a constant is just another numerical constant.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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