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carolinar7

  • one year ago

log(2)6•log(6)8

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  1. Nnesha
    • one year ago
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    is i log(6)2 •log(6)8 base 6 ??

  2. carolinar7
    • one year ago
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    No the first one is base two in the second one is base six

  3. Nnesha
    • one year ago
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    \[\huge\rm log_2 6 \times \log_6 8\] like this

  4. carolinar7
    • one year ago
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    *and

  5. carolinar7
    • one year ago
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    Yes you got it

  6. carolinar7
    • one year ago
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    How do you solve it

  7. carolinar7
    • one year ago
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    @Nnesha

  8. carolinar7
    • one year ago
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    @PhantomCrow

  9. carolinar7
    • one year ago
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    @Michele_Laino

  10. Nnesha
    • one year ago
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    alright familiar the the change of base formula ?

  11. Nnesha
    • one year ago
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    hey ??

  12. Nnesha
    • one year ago
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    huh anywys i have to go chage of base formula \[\huge\rm \log_\color{ReD}{b} \color{blue}{a} =\frac{ \log \color{blue }{a} }{ log\color{ReD}{ b} }\] write both log in fraction ^by using change of base formula

  13. Nnesha
    • one year ago
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    then rewrite 8 in terms of base 2

  14. Nnesha
    • one year ago
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    make sense ?? @carolinar7

  15. Nnesha
    • one year ago
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    i'll do the log(6)8 \[\huge\rm \log_\color{ReD}{6} \color{blue}{8} =\frac{ \log \color{blue }{8} }{ log\color{ReD}{ 6} }\] rewrite 8 in terms of base 2 2 times 2 times 2 = 8 you can write it as 2^3 \[\large\rm \log_\color{ReD}{6} \color{blue}{8} =\frac{ \log \color{blue }{2^3} }{ log\color{ReD}{ 6} }\] apply the power rule power rule \[\large\rm log_b x^y = y \log_b x\] write log(2) 6 in fraction by using change of base formula. that's it

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