Question 179772
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*[tex \Large \log_b {a^n} = n \log_b {a}]


and


*[tex \Large \log_b^{-1}({\log_b {a}}) = a]


and


*[tex \Large \log (a) = \log_{10}(a)]


So: 


*[tex \Large \log^{-1}({\log (1215^{3 \over 2})}) = \log^{-1}({{3 \over 2}\log (1215)}) = 1215^{3 \over 2}]


Using the calculator <u>log</u> (where it says *[tex \Large \log]) and <u> inverse log</u> (where it says (*[tex \Large \log^{-1}]) functions: 


*[tex \Large \log^{-1}({{3 \over 2}\log (1215)}) \approx \log^{-1}({{3 \over 2}(3.0845)}) \approx \log^{-1}({4.6269}) \approx 42351.0729 \approx 1215^{3 \over 2}]


In words, use the log function on the calculator to find the log of 1215.  Multiply this value by 3/2.  Take the inverse log of that result.  Done.


Do the other ones the same way. 


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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