Question 615879
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I have to presume you mean


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4^{2t}\ =\ 5^{4t\,+\,4}]


Because the only way I know to solve


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4^{2t}\ =\ 5^{4t}\ +\ 4]


Is to find an approximate root by numerical methods.


Be that as it may, your intuition is spot on.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4^{2(-2.5803)}\ \neq\ 5^{4(-2.5803)\,+\,4}]


(order of magnitude difference)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4^{2t}\ =\ 5^{4t\,+\,4}]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(4^{2t}\right)\ =\ \ln\left(5^{4t\,+\,4}\right)] 


Use *[tex \LARGE \log_b\left(a^n\right)\ =\ n\log_b(a)]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2t)\ln\left(4\right)\ =\ (4t\,+\,4)\ln\left(5\right)]


Multiply both sides by *[tex \LARGE \frac{1}{2\ln4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ (2t\,+\,2)\frac{\ln\left(5\right)}{\ln\left(4\right)]


Just for neatness sake, let *[tex \LARGE C_1\ =\ \frac{\ln\left(5\right)}{\ln\left(4\right)]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{2C_1}{1\ -\ 2C_1]


The rest is just calculator (or better, spreadsheet) work.


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