## nickersia Group Title Question in comments, can someone help? :) one year ago one year ago

1. nickersia Group Title

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

2. estebananaya Group Title

3. nickersia Group Title

Find x ? :D

4. estebananaya Group Title

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

5. nickersia Group Title

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

6. estebananaya Group Title

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

7. estebananaya Group Title

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

8. nickersia Group Title

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

9. shining Group Title

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

10. shining Group Title

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

11. nickersia Group Title

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

12. estebananaya Group Title

i think both are acceptable

13. nickersia Group Title

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!