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

2 Answers
Dec 3, 2017

x=2^((3+-sqrt5)/2)x=23±52

Explanation:

First we will convert log_x(2)logx(2) to log_2log2. We will use the following log property:
log_b(a)=log_x(a)/log_x(b)logb(a)=logx(a)logx(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)logx(2)+log2(x)=log2(2)log2(x)+log2(x)=1log2(x)+log2(x)

This is a quadratic, but to make it easier to understand I will introduce a substitution so u=log_2(x)u=log2(x):
1/log_2(x)+log_2(x)=31log2(x)+log2(x)=3

1/u+u=31u+u=3

Multiply by uu on both sides:
1+u^2=3u1+u2=3u

u^2-3u+1=0u23u+1=0

Solve using the quadratic formula:
u=(3+-sqrt5)/2u=3±52

Now we resubstitute, so we have
log_2(x)=(3+-sqrt5)/2log2(x)=3±52

We put both sides as powers of 22:
cancel(2)^(cancel(log_2)(x))=2^((3+-sqrt5)/2)

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

Dec 3, 2017

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))