Question 418423
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Presuming you mean:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10^{x\,+\,3}\ =\ 6^{2x}]


Mainly because


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10^{x}\ +\ 3\ =\ 6^{2x}]


can only be solved by numerical methods,


take the base 10 log of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(10^{x\,+\,3})\ =\ \log(6^{2x})]


Use the laws of logarithms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\,+\,3)\log(10)\ =\ 2x\log(6)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 3\ =\ 2x\log(6)]


(Because *[tex \Large \log(10)\ =\ 1])


The rest is elementary algebra:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{3}{2\log(6)\,-\,1}]


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