Question

How do you find the value of x: `log_x (121/289) = 2`?

Answer 1

`x=11/17`

Explanation:

`color(red)((1)log_aX=n<=>X=a^n)`
`color(red)((2)log_aX^n=nlogX)`
We have,
`log_x(121/289)=2=>log_x(11/17)^2=2`
`=>2*log_x(11/17)=2`
`=>log_x(11/17)=1`
`=>11/17=x^1=>x=11/17`