Find the value of $ab$ $$5^a = 16, 8^b = 25$$


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*Find the value of $ab$. 


So, this question seems simple to solve. However, the most important thing is to know where to start. That's why I couldn't solve this problem. I've been looking for a method/strategy to solve all kinda questions which involve exponential terms. 
Now, rewriting the inequalities. 
$$5^a = 16 \implies 5^a = 4^2 \implies 5^a = 2^4$$
$$8^b = 25 \implies 8^b = 5^2  \implies 2^{3b} = 5^2$$
Or what about giving letters like $k$, $t$ or somewhat?
Regards!

$5^a = 16$ so $a =\log_5 16= \log_5 2^4 = 4\log_5 2$
$8^b = 25$ so $b = \log_8 25=\log_8 5^2 = 2\log_8 5$
So $ab =4\log_5 2*2\log_8 5=8\log_5 2*\log_8 5$.  
That may (or may not) be simplified.
Use $\log_a b = \frac {\log_M b}{\log_M a}$ to convert to a common base.
$\log_5 2 = \frac {\ln 2}{\ln 5}$ and $\log_8 5 = \frac {\ln 5}{\ln 8}=\frac {\ln 5}{3\ln 2}$.  (Nothing special about $\ln$.  We could just as easily used $\log_{10}$.  Or $\log$ base anything.)
So $ab = 8\log_5 2*\log_8 5= 8\frac {\ln 2}{\ln 5}\frac {\ln 5}{3\ln 2} = \frac 83$.
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Another trick would be to recognize $\log_a b = \frac 1{\log_b a}$ and that $\log_{a^k} b = \frac 1k \log_a b$. 
So then:
$\log_5 16*\log_8 25 = \frac {\log_5 16}{\log_25 8} = \frac {4\log_5 2}{3*\frac 12 \log_5 2} = \frac 83$
But that a little subtle and sophisticated and it's not reasonable to expect everyone to see that right away.
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I suppose if you get really comfortable with exponents you could simply do:
$8^{ab} = (8^b)^a = 25^a = (5^2)^a = (5^a)^2 = 16^2 = 2^2*8^2$
$ab = \log_8 8^{ab} = \log_8 4*8^2 = \log_8 2^2 + \log_8 8^2 = 2\log_8 2 + 2 = 2\frac 23$.
.... Actually that's pretty slick!
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Actually without logarithms this is okay if you recognize that this are using powers.
$5^a = 2^4$ and $8^b = 2^{3b} = 5^2$.
So we can do either $5^{ab} = 2^{4b}=2^{3b*\frac 43} = (5^2)^{\frac 43}= 4^{\frac 83}$ and $ab = \frac 83$ or we can do $2^{3ab} = 5^{2a} = 2^{8}$ and $ab =\frac 83$.
But that's kind of round about that one simply assumed you knew logarithms:
$a = 4*\log_5 2$ and $3b = 2 \log_2 5$ and so $3ab = 8 \log_5 2\log_2 5$ is the exact same thing.
It's a little be clever to note that $\log_m n *\log_n m = 1$ but if one knows that we can convert bases via $\log_m n = \frac {\log_M n}{\log_M n}$ so $\log_2 5 = \frac {\log_M 5}{\log_M 2}$ and $\log_5 2 = \frac {\log_M 2}{\log_M 2}$ leads to that directly.
