Question 376396
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I presume you mean:


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


Next time use parentheses appropriately, that is: (4/3)^(1 - x) = 5^x would have made your meaning clear.


Take the log of both sides.  Whether you use natural logs or base 10 logs (or indeed any other base) doesn't matter.  I use *[tex \Large \ln], just because it tickles my fancy to do so.


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


Use


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


to write


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


Use ordinary algebra to write:


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


Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b\left(\frac{x}{y}\right)\ =\ \log_b(x) - \log_b(y)]


To write:


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


Use ordinary algebra to write:


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


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


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


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


Use your calculator to find the numeric approximation and to check your answer.


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