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I have stumbled on this question, and there are a few questions after it of the same type. How do I solve it and what is the right approach for this kind of question?

$$8^{2n+1} = 32^{n+1}$$

I need to find the value of $^n$. Step by step if at all possible please. My approach was to start like:

$$(2^3)^{2n+1} = (2^3 \times 2^2)^{n+1}$$

Then I was lost as to how to proceed. Do I need some sort of prerequisite knowledge that I am missing before attempting this? The book has only shown $7$ laws of indices before asking this question.

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The idea of reducing everything to powers of $2$ is good, but you didn’t carry it out correctly on the righthand side: you should have

$$\left(2^3\right)^{2n+1}=\left(2^5\right)^{n+1}\;.\tag{1}$$

Now use the fact that $\left(a^b\right)^c=a^{bc}$ to rewrite $(1)$ as

$$2^{3(2n+1)}=2^{5(n+1)}\;,$$ or $$2^{6n+3}=2^{5n+5}\;.$$

This is the case if and only if $6n+3=5n+5$, so all that remains is to solve that simple linear equation for $n$.

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    $\begingroup$ I almost forgot about $(a^b)^c=a^{bc}$ Practice may make me think of that more quickly. $\endgroup$ – sprocket12 Mar 17 '13 at 21:30
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The general technique is to write both sides with the same base and then equate the powers.

$(2^3)^{2n+1}=(2^5)^{n+1}$

$2^{3(2n+1)}=2^{5(n+1)}$

$3(2n+1)=5(n+1)$

$6n+3=5n+5$

$n=2$

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The other solutions to this problem presuppose two factors:

1) the two bases are themselves both powers of some common base;

2) the solver recognizes that fact.

But suppose that the bases 8 and 32 in the problem above were replaced with 2197 and 371293. Or even worse, with 2197 and 317293.

A more "bullet-proof" technique would be to get the unknown out of the exponents by taking logarithms of both sides. The equation to be solved becomes:$$(2n+1)\log(8)=(n+1)\log (32)$$When you divide log32 by log8 and get $\frac{5}{3}$, it's clear that you missed some subtle point. but you're already well on your way to the solution, and the next time the numbers might not work out so neatly...

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  • $\begingroup$ +1 and probably going from 2^a = 2^b ==> a == b is implicitly taking log_2 of both sides anyway $\endgroup$ – wim Mar 18 '13 at 4:13
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Denote $x=2^n$ and find $x$ out your equation..

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