Question 982472
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^{x+3}\ =\ 3^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(2^{x+3}\right)\ =\ \ln\left(3^x\right)]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x\ +\ 3}{x}\ =\ \frac{\ln(3)}{\ln(2)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ +\ \frac{3}{x}\ =\ \frac{\ln(3)}{\ln(2)}]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{x}\ =\ \frac{\ln(3)\ -\ \ln(2)}{\ln(2)}]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{3\ln(2)}{\ln(3)\ -\ \ln(2)}]




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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \