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

and then solve quadratic

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- Aug 28th 2009, 06:47 AMattwoodpsolve 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

and then solve quadratic - Aug 28th 2009, 06:57 AMynj
- Aug 28th 2009, 08:38 AMDefunkt
You can't do that; what you meant to do is raise both sides to be the exponent of e, and then:

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

This gives $\displaystyle (x-2)(2x-3) = x^2$ , and you know the rest... - Aug 28th 2009, 10:45 AMstapel
Use

**log rules**to convert the equation to:

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

Then equate the arguments, and solve the resulting quadratic equation. (Wink)