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

  • 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!

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  • $\begingroup$ Why don't you use logarithm? $\endgroup$ – Botond Apr 24 '18 at 19:56
  • $\begingroup$ @Botond I don't know how to use logarithm. $\endgroup$ – Displayed Apr 24 '18 at 19:56
  • $\begingroup$ Can't we solve it by giving letters? $\endgroup$ – Displayed Apr 24 '18 at 20:02
  • $\begingroup$ @Displayer You can, as you can see in Delta-u's answer. $\endgroup$ – Botond Apr 24 '18 at 20:04
  • $\begingroup$ @Botond He didn't use any letter. $\endgroup$ – Displayed Apr 24 '18 at 20:11
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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:$$ab=\frac{8}{3}$$

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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 $ab=8/3$.

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  • $\begingroup$ Not quite. $8 \neq 16^{\frac{2}{3}}$. $\endgroup$ – user328442 Apr 24 '18 at 20:01
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    $\begingroup$ Sorry, silly me! I edited. $\endgroup$ – Thomas Bakx Apr 24 '18 at 20:04
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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 $8=2^3$

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  • $\begingroup$ Do you think it is better to use logarithm? Our teacher didn't teach us how to solve these problems with logarithm, that's why. Also, It'd be great if you explain how to solve it with logarithm. $\endgroup$ – Displayed Apr 24 '18 at 20:12
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$5^a = 16$ 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.

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    $\begingroup$ I decided not to use this logarithm anymore when I've seen this answer thanks for breaking my dreams. $\endgroup$ – Displayed Apr 24 '18 at 20:25

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