# `"log"_(5x) (5/x) + ("log"_5 x)^2 = 1` Solve for x

mlehuzzah | Certified Educator

`"log"_(5x)(5/x)+("log"_(5)x)^2=1`

We want to change the base in the first term, so we use the change of base formula to get:

`("log"_5 (5/x))/("log"_5 (5x))+("log"_5 x)^2 = 1` ` `

Now use the fact that `"log" (A/B) = "log" A - "log" B`

and also, `"log" (AB) = "log"A + "log" B`

So now we have:

`("log"_5 5 - "log"_5 x)/("log"_5 5 - "log"_5 x) + ("log"_5 x)^2 = 1`

`"log"_a a = 1` for positive a, so we have:

`(1-"log"_5 x)/(1-"log"_5 x) + ("log"_5 x)^2 = 1`

To make this easier to work with, let `y="log"_5 x`

So:

`(1-y)/(1+y)+y^2 = 1`

Multiply all sides by 1+y (we will have to check later that y is not -1)

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

`y^3 + y^2 - 2y=0`

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

Thus y=0, y=-2, or y=1 (this is good; y is not -1)

`y="log"_5 x` , so `x=5^y`

Thus `x=5^0=1` , `x=5^(-2)=1/(25)` , or `x=5^1=5`

PS. I've edited your question.  I wasn't completely sure where the ^2 should go; hopefully this is what you meant.