Question 628010
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4^{(4x\ -\ 3)}\ =\ 5^{(-x\ -\ 1)}]


Take the log of both sides (any base, doesn't matter)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(4^{(4x\ -\ 3)}\right)\ =\ \ln\left(5^{(-x\ -\ 1)}\right)]


Use *[tex \LARGE \log_b(x^n)\ =\ n\log_b(x)] on both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (4x\ -\ 3)\ln\left(4\right)\ =\ (-x\ -\ 1)\ln\left(5\right)]


Multiply both sides by *[tex \LARGE \frac{1}{\ln(4)}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (4x\ -\ 3)\ =\ (-x\ -\ 1)\frac{\ln\left(5\right)}{\ln\left(4\right)}]


just for the sake of neatness, let *[tex \LARGE k\ =\ \frac{\ln\left(5\right)}{\ln\left(4\right)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (4x\ -\ 3)\ =\ (-x\ -\ 1)k]


Then it is just plain old algebra from here:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x\ +\ kx\ =\ 3\ -\ k]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{3\ -\ k}{4\ +\ k}\ =\ \frac{3\ -\ \frac{\ln\left(5\right)}{\ln\left(4\right)}}{4\ +\ \frac{\ln\left(5\right)}{\ln\left(4\right)}}]


Since you want the exact answer, and you don't care about actually being able to approximate the log values, you can get a much neater answer using base 4 logs:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{3\ -\ \log_4(5)}{4\ +\ \log_4(5)}]


I'll let you figure out why these two answers are the same.  Hint:  What is the value of *[tex \LARGE \log_b(b)]?


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