what did I do wrong? 1+2log_x(5)log_25(2x-5)=1/log_25(x)
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Aug 17, 2008 #2 Re: I'd like to know what i did wrong Hello, oded244! Your last step . . . . . \(\displaystyle \underbrace{\log_{25}(x) + \log_{25}(2x-5)} \;=\;1\) . . . \(\displaystyle \log_{25}[x(2x-5)] \;=\;1 \quad\Rightarrow\quad x(2x-5) \;=\;25^1 \quad\hdots \text{etc.}\)
Re: I'd like to know what i did wrong Hello, oded244! Your last step . . . . . \(\displaystyle \underbrace{\log_{25}(x) + \log_{25}(2x-5)} \;=\;1\) . . . \(\displaystyle \log_{25}[x(2x-5)] \;=\;1 \quad\Rightarrow\quad x(2x-5) \;=\;25^1 \quad\hdots \text{etc.}\)
O oded244 New member Joined Oct 4, 2007 Messages 38 Aug 17, 2008 #3 Re: I'd like to know what i did wrong yea, got it. thanks!
