Solve lnx + ln(x-3) = 0 So far my working is e^(lnx) + e^(ln(x-3)) = e^0 x + (x-3) = 1 and so on But this leads me to the incorrect solution. Where did I go wrong?
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Originally Posted by hymnseeker e^(lnx) + e^(ln(x-3)) = e^0 Where did I go wrong? $\displaystyle \ln(a) + \ln(b) = \ln(ab)$
You can't do what you did. What you can do is $\displaystyle a + b = c \implies e^{a+b} = e^c $
How foolish of me to miss that. Thanks again.
Originally Posted by hymnseeker Solve lnx + ln(x-3) = 0 So far my working is e^(lnx) + e^(ln(x-3)) = e^0 x + (x-3) = 1 and so on But this leads me to the incorrect solution. Where did I go wrong? Hello hymnseeker, Try this: $\displaystyle \ln x + \ln (x-3)=0$ $\displaystyle \ln x(x-3)=0$ $\displaystyle x(x-3)=e^0$ $\displaystyle x^2-3x=1$
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