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anonymous

  • one year ago

What is the value of

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  1. anonymous
    • one year ago
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    \[\log_{81} 3\]

  2. anonymous
    • one year ago
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    @UsukiDoll

  3. anonymous
    • one year ago
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    @DecentNabeel

  4. DecentNabeel
    • one year ago
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    \[\log _{81}\left(3\right)=\frac{1}{4}\quad \left(\mathrm{Decimal:\quad }\:0.25\right)\]

  5. anonymous
    • one year ago
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    @satellite73

  6. DecentNabeel
    • one year ago
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    Happy:) @EllenJaz17

  7. anonymous
    • one year ago
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    how do you value that though? I have to show my work

  8. anonymous
    • one year ago
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    @DecentNabeel

  9. DecentNabeel
    • one year ago
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    \[\mathrm{Rewrite\:}3\mathrm{\:\in\:power-base\:form:}\quad 3=81^{\frac{1}{4}}\] \[=\log _{81}\left(81^{\frac{1}{4}}\right)\] \[\mathrm{Apply\:the\:\log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)\] \[\log _{81}\left(81^{\frac{1}{4}}\right)=\frac{1}{4}\log _{81}\left(81\right)\] \[=\log _{81}\left(81\right)\frac{1}{4}\] \[\mathrm{Apply\:the\:\log\:rule}:\quad \log _a\left(a\right)=1\] \[\log _{81}\left(81\right)=1\] =1/4

  10. DecentNabeel
    • one year ago
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    ok @EllenJaz17

  11. anonymous
    • one year ago
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    \[\log_{81}(3)=x\iff 81^x=3\] so what you are trying to do is figure out how to write 3 as a power of 81 since \(\sqrt[4]{81}=3\) that tells you \[81^{\frac{1}{4}}=3\] and you have the power you are looking for

  12. anonymous
    • one year ago
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    so it's not 1/4 its 3?

  13. anonymous
    • one year ago
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    another easier example \[\log_7(49)=2\] because \(7^2=49\)

  14. anonymous
    • one year ago
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    no the answer is \(\frac{1}{4}\) let me write it again

  15. anonymous
    • one year ago
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    ok sounds good

  16. anonymous
    • one year ago
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    \[\huge \log_{81}(\color{red}x)=3\iff 81^{\color{red}x}=3\] you are trying to find \(\color{red}x\)

  17. anonymous
    • one year ago
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    you are looking for the power that you would raise 81 to to give you 3

  18. anonymous
    • one year ago
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    \[\log_2(8)=3\iff 2^3=8\] they say the same thing

  19. anonymous
    • one year ago
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    you should be able to switch back and forth quickly

  20. anonymous
    • one year ago
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    the only tricky part of this one was recognizing that 3 is the fourth root of 81

  21. anonymous
    • one year ago
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    .sorry catching up was getting something to drink

  22. anonymous
    • one year ago
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    beer i hope

  23. anonymous
    • one year ago
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    haha no just water

  24. anonymous
    • one year ago
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    too bad

  25. anonymous
    • one year ago
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    I prefer straight tequila, gets the job done quick lol

  26. anonymous
    • one year ago
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    no beer chaser? dang!

  27. anonymous
    • one year ago
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    Haha nope!

  28. anonymous
    • one year ago
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    oops do you get that \(2^5=32\) says the same thing as \(\log_2(32)=5\) ?

  29. anonymous
    • one year ago
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    how would i apply that to find this value in my question?

  30. anonymous
    • one year ago
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    so if you were trying to solve \[\log_2(32)=\color{red}x\] you would be trying to solve \[2^{\color{red}x}=32\]

  31. anonymous
    • one year ago
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    you have \[\log_{81}(3)=\color{red}x\] you have to solve \[81^{\color{red}x}=3\]

  32. anonymous
    • one year ago
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    since \(\sqrt[4]{81}=x\) that is the same think as saying \(81^{\frac{1}{4}}=3\)

  33. anonymous
    • one year ago
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    ooops i means "since \(\sqrt[4]{81}=3\)

  34. anonymous
    • one year ago
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    now you have the power you are looking for , that power is \(\frac{1}{4}\) therefore since \[81^{\frac{1}{4}}=3\] it is the same as saying \[\log_{81}(3)=\frac{1}{4}\]

  35. anonymous
    • one year ago
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    Ok i kinda get it lol

  36. anonymous
    • one year ago
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    yeah it is confusing at first a simpler example is it clear that \(10^4=10,000\)?

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