Thread: 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

and then solve quadratic

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

and then solve quadratic

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

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

and then solve quadratic

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



This gives , and you know the rest...

Originally Posted by attwoodp

ln(x-2)+ln(2x-3)=2lnx solve for x.

Use log rules to convert the equation to:



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