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log base5 2+2 log base5 x=log base5 (3-x)
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\[\log_{5}2+2\log_{5}x=\log_{5}(3-x) \]
You could use log_5(2) where we'll understand that you mean \(\log_52\)
Convert the left hand side into one thing first
Would they cancel out and I would be left with just x?
I don't exactly get what you mean by cancel out, but you'd need the log formulas for this: \[\Large\log_na+\log_nb=\log_n(ab)\]\[\Large n\log_ma=\log_ma^n\]
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Okay one sec lemme try to figure it out :)
okay :)
Kinda stuck :/
Using the second formula, \(\Large2\log_5x=?\)
\[\log_{5}x^2 \] ?
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Yep
Oh my . I had that written down and I thought it was wrong -_-
Okay thanks
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