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nickersia

  • 2 years ago

Question in comments, can someone help? :)

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  1. nickersia
    • 2 years ago
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    \[4*\log_{x}2=\log_{2}x+3 \] (4*logx 2 = log2 x + 3)

  2. estebananaya
    • 2 years ago
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    what is your question, exactly?

  3. nickersia
    • 2 years ago
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    Find x ? :D

  4. estebananaya
    • 2 years ago
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    i think you can start by putting logx 2=log2 2/ log2 x

  5. nickersia
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    i didn't get the answer, but i think the procedure is correct

  8. nickersia
    • 2 years ago
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    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
    • 2 years ago
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    first write log 2 x into 1/(logx 2) then t=logx 2 solve it!

  10. shining
    • 2 years ago
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    hint:you will get two values that just one of them is acceptable

  11. nickersia
    • 2 years ago
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    I got for x 2 and 1/16, why is only one acceptable?

  12. estebananaya
    • 2 years ago
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    i think both are acceptable

  13. nickersia
    • 2 years ago
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    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!

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