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

##### 2 Answers

#### Explanation:

First we will convert

In our equation, this becomes:

This is a quadratic, but to make it easier to understand I will introduce a substitution so

Multiply by

Solve using the quadratic formula:

Now we resubstitute, so we have

We put both sides as powers of

#### 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

#1/t + t = 3#

Mutliplying through by

#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))#