What is the answer to $ \log_5 (25\sqrt 5)$?

Question

I have started to do this question, but not sure if I am on the right track or not (or even allowed in math)

Can someone say if what I am doing is right?

Answer 1

Alternatively:

$25= 5^2$ and $\sqrt 5 = 5^{\frac 12}$ and

$25\sqrt{5} = 5^2*5^{\frac 12} = 5^{2+\frac 12}=5^{\frac 52}$.

There are many ways to interpret this: $(5^{\frac 52} = \sqrt{5^5}=\sqrt{3125}=25\sqrt 5$ or $5^{\frac 52}=(\sqrt 5)^5 = (\sqrt 5)^2*(\sqrt 5)^2*\sqrt 5 = 5*5*\sqrt 5 = 25\sqrt 5$.

So all of this should make sense.

So $\log_5 25\sqrt 5 = \log_5 5^{\frac 52} = \frac 52$.

Answer 2

Yep your solution is correct except for the fact that you have multiplied when opening the logarithm. This is the correct identity [log(ab)]=loga+logb and not loga.logb