Question

Solve for X, log X to the base 3 + log √x to the base 3 + log 4 th √x to the base 3 + log 8th√x to the base 3+........=4 , (X >0)

Answer 1

We have,

$\log_3x+\dfrac{1}{2}\log_3x+\dfrac{1}{4}\log_3x+\dfrac{1}{8}\log_3x+\dots=4$

by the rule,

$\log_m\left(a^b\right)=b\log_m(a)$

So,

$\log_3x+\dfrac{1}{2}\log_3x+\dfrac{1}{4}\log_3x+\dfrac{1}{8}\log_3x+\dots=4\\\\\\\log_3x\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dots\right)=4\\\\\\\log_3x\left(\dfrac{1}{\frac{1}{2}}\right)=4$

This is because, since $1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dots$ is an infinite geometric series, we know,

$\displaystyle\sum_{k=1}^{\infty}ar^{k-1}=\dfrac{a}{1-r}$ for $|r|<1.$

So,

$\log_3x\left(\dfrac{1}{\frac{1}{2}}\right)=4\\\\\\\log_3x=2\\\\\\\implies\ \large\boxed{\mathbf{x=9}}$