find x

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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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find x ^ | | | here it is lol
we just found x yesterday
today we should find new and exciting things

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Other answers:

Equation?
\[(2x)^{\log_{2} }+(3x)^{\log_{3} }=0\]
------------------> x <-------------------------
exp are exp
your logs have no arguemtns
is that spose to be: log2(2x) + log3(3x) = 0 by chance?
just solve it
just notate it ....
so far youve written something thats just jibberish
you need arguments after the logs, you have a typo
i know log2(2x) + log3(3x) = 0
you guys are funny XD
ok, that I can do :) maybe log2(2)+log2(x)+log3(3)+log3(x) = 0 log2(x)+log3(x)= -2 hmmm
Do you mean |dw:1327678126892:dw|
change of base :) \[\frac{ln(x)}{ln(2)}+\frac{ln(x)}{ln(3)}=-2\]
we would need a common denominator with this i assume
\[\frac{ln(3)ln(x)}{ln(3)ln(2)}+\frac{ln(2)ln(x)}{ln(2)ln(3)}=\frac{-2}{ln(2)ln(3)}\]
equate the tops\[ln(3)ln(x)+ln(2)ln(x)=-2\] factor out the ln(x) \[ln(x)(ln(3)+ln(2))=-2\] divide off the not ln(x) \[ln(x)=-\frac{2}{(ln(3)+ln(2))}\] and e both sides
\[x=exp(-\frac{2}{ln(3)+ln(2)})\] and that is my gut effort
http://www.wolframalpha.com/input/?i=log_3%283x%29%2Blog_2%282x%29%3D0 forgot to include the "common denominator" in the top, but not bad
\[x=exp(-2\frac{ln(2)ln(3)}{ln(3)+ln(2)})\]

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