Question 984421
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Use the definition of the log function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(3)\ =\ y_1\ \Right\ 2^{y_1}\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_{\frac{1}{2}}(5)\ =\ y_{2}\ \Right\ \left(\frac{1}{2}\right)^{y_{2}}\ =\ 5]


Clearly *[tex \Large y_1] is some small but positive number.  But if you raise a number between zero and 1 to a positive power, you get a smaller number.  Hence, *[tex \Large y_2] must be a negative number, and is therefore the smaller value.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \