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nickersiaBest ResponseYou've already chosen the best response.0
\[4*\log_{x}2=\log_{2}x+3 \] (4*logx 2 = log2 x + 3)
 one year ago

estebananayaBest ResponseYou've already chosen the best response.1
what is your question, exactly?
 one year ago

estebananayaBest ResponseYou've already chosen the best response.1
i think you can start by putting logx 2=log2 2/ log2 x
 one year ago

nickersiaBest ResponseYou've already chosen the best response.0
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
 one year ago

estebananayaBest ResponseYou've already chosen the best response.1
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
 one year ago

estebananayaBest ResponseYou've already chosen the best response.1
i didn't get the answer, but i think the procedure is correct
 one year ago

nickersiaBest ResponseYou've already chosen the best response.0
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
 one year ago

shiningBest ResponseYou've already chosen the best response.0
first write log 2 x into 1/(logx 2) then t=logx 2 solve it!
 one year ago

shiningBest ResponseYou've already chosen the best response.0
hint:you will get two values that just one of them is acceptable
 one year ago

nickersiaBest ResponseYou've already chosen the best response.0
I got for x 2 and 1/16, why is only one acceptable?
 one year ago

estebananayaBest ResponseYou've already chosen the best response.1
i think both are acceptable
 one year ago

nickersiaBest ResponseYou've already chosen the best response.0
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!
 one year ago
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