# Math Help - Solving equation involving e and ln

1. ## Solving equation involving e and ln

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?

2. Originally Posted by hymnseeker

e^(lnx) + e^(ln(x-3)) = e^0

Where did I go wrong?
$\ln(a) + \ln(b) = \ln(ab)$

3. You can't do what you did.
What you can do is $a + b = c \implies e^{a+b} = e^c$

4. How foolish of me to miss that.

Thanks again.

5. 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:

$\ln x + \ln (x-3)=0$

$\ln x(x-3)=0$

$x(x-3)=e^0$

$x^2-3x=1$