Question

Solve the equation `25log_2x=log_x2` ?

`25log_2x=log_x2`
I have gotten to `logx^25=1/(logx)` but I am not sure how to proceed from here. I am supposed to solve this without the use of a calculator, and the answer is given as `x=2^((+/-)1/5)`. How should I go about solving this?

Answer 1

I tried this:

Explanation:

I would try changing the base of the second log as:

`log_x2=log_2(2)/log_2(x)=1/log_2(x)`

so we get:

`25log_2x=1/log_2(x)`

rearrange:
`25log_2x*log_2x=1`
`25[log_2x]^2=1`
`[log_2x]^2=1/25`
take the square root of both sides to get rid of the square:
`log_2x=+-1/5`
apply the definition of log:
`x=2^(+-1/5)`

Answer 2

`x=1/2^(1/5)`, `x=2^(1/5)`

Explanation:

`25log_2(x)=log_x(2)`
`log_2(x^(25))=log_x(2)`
`log(x^(25))/log(2)=log(2)/log(x)`
`25log(x)*log(x)=log(2)*log(2)`
`25log^2(x)=log^2(2)`
`5log(x)=+-log(2)`
`log(x^5)=+-log(2)`

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

`log(x^5)=log(2)`
`x^5=2`
`x=2^(1/5)`