Find the value of $ab$

Question

$$5^a = 16, 8^b = 25$$

• 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!

Answer 1

Notice that: $$8^{ab}=\left(8^b \right)^a=25^a=5^{2a}=\left(5^a \right)^2=16^2$$ So: $$2^{3ab}=2^8$$ and:$$ab=\frac{8}{3}$$

Answer 2

Note that $5^a = 16$ so that $5^{\frac{3a}{4}}=8$ since $16^3 = 8^4$. But then $8^b = 5^{\frac{3ab}{4}} = 25$, so that $\frac{3ab}{4} = 2$, or $ab=8/3$.

Answer 3

The standard way: $$5^a=16\implies a=\frac{\log16}{\log5}$$ and $$8^b=25\implies b=\frac{\log25}{\log8}$$ so $$ab=\frac{\log16}{\log8}\cdot\frac{\log25}{\log5}=\frac{\log{8}+\log{2}}{\log8}\cdot2=\left(1+\frac{1}{3}\right)\cdot2=\frac83$$ since $8=2^3$

Answer 4

$5^a = 16$ 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 $ab = 8\log_5 2*\log_8 5= 8\frac {\ln 2}{\ln 5}\frac {\ln 5}{3\ln 2} = \frac 83$.

...

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.

....

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!

....

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.