1. ## solve logarithmic equation

ln(x-2)+ln(2x-3)=2lnx

solve for x. My brain must have gone to sleep, i have the answer but can't understand why i can't use:

e^ln(x-2) + e^ln(2x-3) = e^lnx^2

(x-2) + (2x-3) = x^^2

2. Originally Posted by attwoodp
ln(x-2)+ln(2x-3)=2lnx

solve for x. My brain must have gone to sleep, i have the answer but can't understand why i can't use:

e^ln(x-2) + e^ln(2x-3) = e^lnx^2

(x-2) + (2x-3) = x^^2

e^ln(x-2) + e^ln(2x-3) = e^lnx^2

(x-2) + (2x-3) = x^^2
it should be
e^ln(x-2) * e^ln(2x-3) = e^lnx^2

(x-2)* (2x-3) = x^^2

3. Originally Posted by attwoodp
ln(x-2)+ln(2x-3)=2lnx

solve for x. My brain must have gone to sleep, i have the answer but can't understand why i can't use:

e^ln(x-2) + e^ln(2x-3) = e^lnx^2

(x-2) + (2x-3) = x^^2

You can't do that; what you meant to do is raise both sides to be the exponent of e, and then:

$e^{ln(x-2)+ln(2x-3)} = e^{2lnx} \Rightarrow e^{ln( (x-2)(2x-3) )} = e^{ln (x^2)}$

This gives $(x-2)(2x-3) = x^2$ , and you know the rest...

4. Originally Posted by attwoodp
ln(x-2)+ln(2x-3)=2lnx solve for x.
Use log rules to convert the equation to:

$\ln(2x^2\, -\, 7x\, +\, 6)\, =\, \ln(x^2)$

Then equate the arguments, and solve the resulting quadratic equation.