We have `log_((log_(5)x))5=2`
Base of logarithm `gt0` and `!=1`
`:.log_(4)xgt0` and `log_(5)x!=1`
`impliesxgt1` and `x!=5`
`:.` The original equation is equivalent to
`log_(5)x=5^(1//2)=sqrt(5)` ltrgt `:.x_(1)=5^(sqrt(5))`
Here `5^(sqrt(5))` is the only root of the original equation.
