Question 1171086
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Since *[tex \Large 25\ =\ 5^2], *[tex \Large \log_a(25)\ =\ \log_a\(5^2\)]


So *[tex \Large \log_a(25)\ =\ 2\log_a(5)\ =\ 1.1292]


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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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