Question in comments, can someone help? :)

Question

anonymous · Answer

$$4*\log_{x}2=\log_{2}x+3  $$
(4*logx 2 = log2 x + 3)

anonymous · Answer

what is your question, exactly?

anonymous · Answer

Find x ? :D

anonymous · Answer

i think you can start by putting logx 2=log2 2/ log2 x

anonymous · Answer

The hint we get is change the base...
There's formula
$$\log_{a} b=\frac{ \log_{c}b  }{ \log_{c}a  }$$
But I tried, I can't solve it

anonymous · Answer

you got then 4log2 2=(log2 x)*(log2 x + 3)
and finally: (log2 x)^2+3log2 x + 4log2 2=0
now i think you can substitute log2 x by u, solve the quadratic equation and then once you have the values of u, compute the values for x:
log2 x=u    ----->  x=2^u

anonymous · Answer

i didn't get the answer, but i think the procedure is correct

anonymous · Answer

Oh, I think I get the idea, I'll try to solve it now, thank you!
That was exactly what I was doing but I didn't saw that I can change log2 x with u and get quadratic

anonymous · Answer

first write log 2 x into 1/(logx 2)
then t=logx 2
solve it!

anonymous · Answer

hint:you will get two values that just one of them is acceptable

anonymous · Answer

I got for x 2 and 1/16, why is only one acceptable?

anonymous · Answer

i think both are acceptable

anonymous · Answer

So and I, thanks guys, you helped me a lot :)
It was important to me to get the idea how to do it, cause exam is tomorrow, and that was the only one I was struggling :) Thank you, all the best!