Stupid question I know but how would I solve this... log x ^ log x = 25?

Answer 1

You haven't put brackets in, so the question looks ambiguous.

It could be [log(x)] ^ [log(x)], which is difficult,
or it could be log[x ^ log(x)], which is easier.

I'll take the easy way out.

Let log[x ^ log(x)] = 25, where I'll assume log is to the base 10.

By one of the rules of logarithms, which states that
log(M ^ N) = N * log(M), we have :

log(x) * log(x) = 25
or, [log(x)] ^ 2 = 25

Taking the square root of both sides gives :
log(x) = ± 5

As I've assumed this is to base 10, this gives :
x = 10 ^ (± 5)
So, x = 10 ^ 5 or 10 ^ (-5),
which is 100,000 or 0.00001.

Answer 2

log x ^ log x = 25
let f(x) = log x ^ log x -25

we have to solve x for f(x) = 0

this can be done by Newton Raphson method

Answer 3

lg (x ^ lg x) = 25
(lg x) (lg x) = 25
(lg x) ^ 2 = 25
lg x = +5 or - 5

when lg x is +5 , x = 100000
when lg x is -5 , x = 0.00001

Answer 4

answer=log100,000^log100=25