Hello every body! How can i solve this problem: $\displaystyle 1+\frac{log_3(12-X)}{log_3X}=\frac{3}{log_3X} $ Thank you!
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Start by multiplying both sides through by $\displaystyle \log_3x$ giving $\displaystyle \log_3x+\log_3(12-x)=3$ now use the rules $\displaystyle \log_ac+\log_ac = \log_a(b\times c)$ and $\displaystyle \log_aa=1$
Thank you very much!
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