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How do I solve $2^{x-1}=3^{x+a}$? I cannot solve it and have spent an hour on it trying many different ways. Please help me! Thank you!

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up vote 6 down vote accepted

To get the variables in a manageable place, take the logarithm of both sides of the equation (it does not matter what base you use; the power law for logarithms will let you bring the powers in front of the logarithm): $$ 2^{x-1}=3^{x+a}\iff\ln 2^{x-1}=\ln 3^{x+a}\iff (x-1)\ln 2= (x+a)\ln 3 $$

The above is valid, since logarithm functions are one-to-one, and since the first equation above has positive quantities on both sides.

Generally, if you have an equation with the variable appearing in an exponent, you can try (perhaps after a bit of algebra) taking logarithms to produce a more manageable equation as in the case above.

Finishing this problem: $$\eqalign{ &(x-1)\ln 2= (x+a)\ln 3\cr \iff& x\ln 2-\ln 2 = x\ln 3+a\ln 3\cr \iff &x\ln 2-x\ln 3= \ln 2+a\ln 3\cr \iff &x (\ln 2- \ln 3)= \ln 2+a\ln 3\cr \iff &x = {\ln 2+a\ln 3\over \ln 2- \ln 3 }\cr &= { \ln(2\cdot 3^a)\over \ln(2/3)}. } $$

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+1. In fact ${ \log(2\cdot 3^a)\over \log(2/3)}$ to any base –  Henry Dec 24 '11 at 12:52
    
Thank you so much David! ^^ –  tina nyaa Dec 25 '11 at 1:12
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