Solving for x in logarithm problem

[Sumedh]

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.

 

[SammyS]

Show what you tried so far.

What results did you have?

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

 

[Sumedh]

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.

 

[Yuqing]

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.

 

[Sumedh]

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

 

[Harrisonized]

log_x^2 16 + log_2x 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

Solve the quadratic formula for log x and you'll get your answer.

 

[Sumedh]

Thanks a lot i have got the answer
answer are x=4 and x=1/\sqrt[3]{}2