Question

How do you solve `log_x 2 + log_2 x = 3` ?

Answer 1

`x=2^((3+-sqrt5)/2)`

Explanation:

First we will convert `log_x(2)` to `log_2`. We will use the following log property:
`log_b(a)=log_x(a)/log_x(b)`

In our equation, this becomes:
`log_x(2)+log_2(x)=log_2(2)/log_2(x)+log_2(x)=1/log_2(x)+log_2(x)`

This is a quadratic, but to make it easier to understand I will introduce a substitution so `u=log_2(x)`:
`1/log_2(x)+log_2(x)=3`

`1/u+u=3`

Multiply by `u` on both sides:
`1+u^2=3u`

`u^2-3u+1=0`

Solve using the quadratic formula:
`u=(3+-sqrt5)/2`

Now we resubstitute, so we have
`log_2(x)=(3+-sqrt5)/2`

We put both sides as powers of `2`:
`cancel(2)^(cancel(log_2)(x))=2^((3+-sqrt5)/2)`

`x=2^((3+-sqrt5)/2)`

Answer 2

`x = 2^(1/2(3+-sqrt(5))`

Explanation:

Given:

`log_x 2 + log_2 x = 3`

By the change of base formula, we have:

`log 2 / log x + log x / log 2 = 3`

Letting `t = log x / log 2 = log_2 x`, that becomes:

`1/t + t = 3`

Mutliplying through by `t` and rearranging a bit:

`0 = 4(t^2-3t+1)`

`color(white)(0) = 4t^2-12t+4`

`color(white)(0) = (2t)^2-2(2t)(3)+3^2-5`

`color(white)(0) = (2t-3)^2-(sqrt(5))^2`

`color(white)(0) = ((2t-3)-sqrt(5))((2t-3)+sqrt(5))`

`color(white)(0) = (2t-3-sqrt(5))(2t-3+sqrt(5))`

So:

`2t = 3+-sqrt(5)`

That is:

`2log_2 x = 3+-sqrt(5)`

So:

`x = 2^(1/2(3+-sqrt(5))`