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

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