# Solving for x in logarithm problem

## Homework Statement

The equation (attached as image) has
(a)one irrational solution
(b)no prime solution
(c)two real solutions
(d)one integral solution

i would like to get help on how to find the possible values of x

## Homework Equations

( the equation is attached below)

## The Attempt at a Solution

i solved the equation by applying the base changing property and then doing
L.C.M of both equations but after that i am unable to solve further?

Any help will be highly appreciated.

#### Attachments

• equation.png
935 bytes · Views: 518

SammyS
Staff Emeritus
Homework Helper
Gold Member
Show what you tried so far.

What results did you have?

I suggest: Do change of base with base of 2 .

The equation (attached as image) has
(a)one irrational solution
(b)no prime solution
(c)two real solutions
(d)one integral solution

i would like to get help on how to find the possible values of x

i solved the equation by applying the base changing property and then doing
L.C.M of both equations but after that i am unable to solve further?

Any help will be highly appreciated.

#### Attachments

• equatin.png
920 bytes · Views: 432
Because the numbers seem to be related to powers of 2, I was convinced to work with logarithm base 2. Let $lb$ stand for $log_{2}$.

We then have

$$log_{x^2}(16) + log_{2x}(64)$$
$$= \frac{lb(16)}{2\cdot lb(x)} + \frac{lb(64)}{lb(2x)}$$
$$= \frac{2}{lb(x)} + \frac{6}{lb(x) + 1} = 3$$

this reduces to a quadratic which should be easily solvable.

2/log2(x) + 6/ (log2(x)+1)=3
i got this result
after that what should i do

logx^2 16 + log2x 64 = 3
log(16) /log (x2) + log(64) /log(2x) = 3
log(2x) log(16) + log(64) log (x2) = 3 log(2x) log (x2)
log(2x) log(24) + log(26) log (x2) = 3 log(2x) log (x2)
4 (log 2+log x) log(2) + 12 log(2) log (x) = 6 (log 2 + log x) log x
4 (log 2)2 + 4 log (x) log (2) + 12 log (2) log (x) = 6 log (2) log (x) + 6 (log x)2
4 (log 2)2 + 10 log (2) log (x) = 6 (log x)2

answer are x=4 and x=1$/\sqrt[3]{}2$