# x^(2logx)-10x=0 (2logx) is  the exponent

mlehuzzah | Certified Educator

`x^(2 "log" x) - 10x = 0`

`x^(2 "log" x) = 10x`

Take the log of both sides:

`"log" ( x^(2 "log" x) ) = "log" (10x) `

On the left hand side, use the fact that `"log" (A^B) = B "log" A`

On the right hand side, use the fact that `"log" (AB) = "log" A + "log" B`

`2 "log" x "log" x = "log" 10 + "log" x`

Set `y= "log" x`

`2y^2 = "log" 10 + y`

`2y^2 - y - "log" 10 = 0`

This is a quadratic, so we may factor, or, if that doesn't work, we may use the quadratic formula.

By the way, usually in high school textbooks, if you have a "log" and no base is given, the base is 10. (And ln is used for base e). In some college text books, if no base is given, the assumption is base e, even if "log" was used instead of "ln." When I turn log 10 into a number, I'm assuming base 10.

`2y^2 - y - 1 = 0.

This factors to:

`(2y+1)(y-1)=0`

So: `2y+1 = 0` or `y-1=0`

So: `y = -1/2` or `y=1`

`y= "log" x`, so `10^y = x`

So `x=10^(-1/2)` or `x=10^1`

So `x=.31623` or `x=10`