Question 183555
I'll do the first two to get you started


# 1


{{{log(2,(32))}}} Start with the given logarithm



{{{log(2,(2^5))}}} Rewrite 32 as {{{2^5}}}. The key here is we want the base to be 2



{{{5*log(2,(2))}}} Use the identity {{{log(b,(x^y))=y*log(b,(x))}}} to rewrite the expression.



{{{5*1}}} Evaluate the log base 2 of 2 to get 1



{{{5}}} Multiply



So {{{log(2,(32))=5}}}



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# 2

{{{log(10,(-100))}}} Start with the given logarithm



{{{y=log(10,(-100))}}} Set "y" equal to the given logarithm



{{{10^y=-100}}} Convert to exponential form.



Now remember, any positive number raised to a variable power will ALWAYS be positive (ie {{{a^x>0}}} where "a" is a constant). 


So {{{10^y}}} is ALWAYS positive. So {{{10^y=-100}}} is NOT true for any "y" value.



So there are no solutions