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I am trying to figure out how to simplify adding of 2 log terms with the same base but different power.

For example in my textbook:

\begin{align} \ln(x) &= 0.8\ln\left(20000\left(1+\frac r{100}\right)\right) + 0.2\ln\left(20000\left(1-\frac s{100}\right)\right)\\ \ln(x) &= 0.8\left[\ln(20000) + \ln\left(1+\frac r{100}\right)\right] + 0.2\left[(\ln(20000) + \ln\left(1-\frac s{100}\right)\right]\\ \ln(x) &= \ln(20000) + 0.8\ln\left(1+\frac r{100}\right) + 0.2\ln\left(1-\frac s{100}\right) \end{align}

This is what I have got so far

The final simplification is:

$$x = 20000\left(1+\frac r{100}\right)^{0.8}\left(1-\frac s{100}\right)^{0.2}$$

Any help will be greatly appreciated. Thank you.

Another User
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murphyHL
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1 Answers1

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Hope I have understood your question correctly:

Using rules of logs you can put 0.8 and 0.2 back into the “power”. (E.g. 2log(3)=log(3^2))

Then you get ln(..^0.8) + ln(..^0.2)

Then using log rule for addition we get: Ln(x) = Ln(..^0.8* ..^0.2)

Then you can ignore the logs to get to the final simplification.

bananapeel22
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