Given the exponential equation 2^x = 128, what is the logarithmic form of the equation in base 10?

Question

anonymous · Answer

$$\begin{align*}2^x&=128\
\log_{10}2^x&=\log_{10}128\
x\log_{10}2&=\log_{10}{128}\
x&=\frac{\log_{10}128}{\log_{10}2}\
x&=\frac{\log_{10}2^7}{\log_{10}2}\
x&=\frac{7\log_{10}2}{\log_{10}2}\
x&=7
\end{align*}$$

mathmale · Answer

I'd suggest you just apply the "log" operator to both sides of this equation.

Next, look at that log 128.  Can that 128 be factored?  If so, identify its factors and find the sum of the logs of those factors.  See anything in common between the left and right sides of this equation?  

SithsAndGiggles is using the Change of Base formula, which is a very effective approach to this kind of problem.

mallorysipp234 · Answer

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mallorysipp234 · Answer

it would be D, right? @mathmale @SithsAndGiggles