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$$3^{1 + 2\log_3(y-x)} = 48$$ With this problem I have difficulty getting rid of the exponent.

$2\log_5(2y - x - 12) = \log_5(y-x) + \log_5(y + x)$

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    I'm confused about what you tried to do. Did you try to take the log base $5$ of both sides? If so, why base $5$? The natural base of logarithm to use is $3$, as that's the base of the exponential in the problem.2012-11-01
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    As it is pointed out below, you don't even need to do this, as $3^{2 \log_{3}(y-x)}=3^{\log_{3}(x-y)^2}=(x-y)^2$2012-11-01

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