How do you condense #2 Log5 x - log5 y #?

1 Answer
reudhreghs
Apr 11, 2016

#log(x^2/y)#

Explanation:

Condense the first #log#, using the law that a coefficient outside a logarithm is the same as a power inside.

#2log5x = log5x^2#

Now condense everything, knowing that #loga - logb = log(a/b)#.

#log5x^2 - log5y = log((5x^2)/(5y)) = log(x^2/y)#

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