log(2)6•log(6)8

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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is i log(6)2 •log(6)8 base 6 ??
No the first one is base two in the second one is base six
\[\huge\rm log_2 6 \times \log_6 8\] like this

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*and
Yes you got it
How do you solve it
alright familiar the the change of base formula ?
hey ??
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
then rewrite 8 in terms of base 2
make sense ?? @carolinar7
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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