Logarithms and ratios.

Question

This is the question: $$\log_b 64 = \frac{3}{b}$$

And have to find $b$.

So I tried a bit and got this:$$\frac{b}{\log b} = \frac{\log 64}{3}$$

But have no idea what to do next.

Thanks for your help.

Answer 1

$$\log _b 4^{3} = 3 \log _b 4 = \frac{3}{b} \implies \log _b 4 = \frac{1}{b} \implies b^{\frac{1}{b}} = 4$$

Edit:

$$\begin{align} b = 4 ^b \implies \ln b = b \ln 4 &\implies b = e^{b \ln 4} \implies be^{-b\ln 4} = 1 \\&\implies -b\ln 4 e^{-b\ln 4} = -\ln 4 \implies b = \color{red}{-\frac{W (-\ln 4)}{\ln 4 }} \end{align}$$

where $W$ is the Lambert W function.

Answer 2

If anyone can edit this for me , it would be helpful :)

$$\log _b 4^{3} = 3 \log _b 4 = \frac{3}{b} \implies \log _b 4 = \frac{1}{b}$$

So:

$$\frac{\log b}{b} = \log 4$$

Using equation:

if log to base b (n) = x

then b^x = n

hence we can say:

$b = 10^{\log 4}$

$b = 4$

NOTE: All the log with no base mentioned has a base 10