Logarithm!

Prove from it.

$\log_a{B}=C$

$\large{a^c=B}$

$\large{a^4=81}$

Get $\large4^{th}$ root of $\large{81}$ .

I do it by factoring into prime factor.

$\large step{1}$

$\large{81}$ can't divided by $\large{2}$ .

$\large step{2}$

$\large{81÷3=27}$

$\large step{3}$

$\large{27÷3=9}$

$\large step{4}$

$\large{9÷3=3}$

$\large step{4}$

$\large{3÷3=1}$

It's mean's that the $\large 4^{th}$ root of $\large{81}$ is $\large {3}$ .

So is correct $\large {3^4=81}$ .

Hence the Answer is $\color{#3D99F6}{3}$ .

$\large{Note=}$ We use the rooting in logarithm when $\large{a^c=B}$ realy equal No that $\large{a^c}$ $\leq{B}$ .

$\begin{aligned} \log_a 81 &= 4 \\ 81 & = a^4 \\ \end{aligned}$

$\implies a = \boxed{3}$

$log_a81=4$

$\Updownarrow$

$\frac{ln 81}{ln a}=4$

$\frac{ln81}{4}=lna$

$a = e^{\frac{ln81}{4} } =3$