Rewriting 4^(n) as a^(n-1)

Question

How can I rewrite $4^n$ as $a^{n-1}$? Where $a$ is known?

I thought about solving with logarithms but that seems long-winded for something so simple... I think I am just too tired right now, I cannot think.

$$ 4^n=a^{n-1} \implies n \log 4 = (n-1) \log a \\ \frac{n\log4}{n-1}=\log a\implies a=4^{\frac{n}{n-1}}$$

Is that correct? There must be a better way

score 3 · Accepted Answer · answered Dec 09 '16 at 01:44

3

We write $$4^n=a^{n-1}\implies 4\times 4^{n-1}=a^{n-1}\implies 4=\left(\frac a4 \right)^{n-1}\implies \frac a4=\sqrt[{n-1}]4\implies a= 4\times \sqrt[{n-1}]4$$

answered Dec 09 '16 at 01:44

lulu

70,402

Does this mean $a^{n-1}=4^{n-1}\cdot 4$? – minseong Dec 09 '16 at 01:51
Of course. The right hand is $4^{n-1+1}=4^n$ so this is just the relation you started with. – lulu Dec 09 '16 at 02:05

Rewriting 4^(n) as a^(n-1)

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