Question 763938
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You really need to use parentheses to group things together.  I have to assume that the *[tex \LARGE y] in the LHS is part of the exponent on 2 since the problem isn't solvable algebraically otherwise.  But for the sake of clear understanding you should have posted


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ]2^(x+y) = 6^y


Going on that presumption


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^x\,\cdot\,2^y\ =\ 3^y\,\cdot\,2^y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^x\ =\ 3^y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\log_b(2)\ =\ y\log_b(3)]


Note:  The base you choose for the logarithm doesn't matter -- *[tex \LARGE \frac{\log_{b_1}\left(x_1\right)}{\log_{b_1}\left(x_2\right)}\ =\ \frac{\log_{b_2}\left(x_1\right)}{\log_{b_2}\left(x_2\right)}\ \forall b_i\ \in\ \mathbb{R}\ >\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x}{y}\ =\ \frac{\log_b(3)}{\log_b(2)}]



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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