How do you solve log_x 2 + log_2 x = 3logx2+log2x=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))