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

estebananaya
 2 years ago
Best ResponseYou've already chosen the best response.1what is your question, exactly?

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

nickersia
 2 years ago
Best ResponseYou've already chosen the best response.0The 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

estebananaya
 2 years ago
Best ResponseYou've already chosen the best response.1you 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

estebananaya
 2 years ago
Best ResponseYou've already chosen the best response.1i didn't get the answer, but i think the procedure is correct

nickersia
 2 years ago
Best ResponseYou've already chosen the best response.0Oh, 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

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

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

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

estebananaya
 2 years ago
Best ResponseYou've already chosen the best response.1i think both are acceptable

nickersia
 2 years ago
Best ResponseYou've already chosen the best response.0So 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!
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