Question 751134
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^{x\,+\,1}\ -\ 6^{x}\ =\ 2^{x\,-\,1}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\,\cdot\,2^x\ -\ 2^x\,\cdot\,3^x\ =\ \frac{1}{2}\,\cdot\,2^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ -\ 3^x\ =\ \frac{1}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3^x\ =\ \frac{3}{2}]


Take the log of both sides, any base...


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


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


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


Which is the exact answer.  Use your calculator for a numerica approximation.  If you don't care about the numeric approximation, you can use base 3 logs so that your denominator becomes 1.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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