Question

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

Solve for x

Answer 1

I divide both sides with ln?

Answer 2

Some formulas for you:
\[\ln a+\ln b = \ln ab\\
n\ln a = \ln a^n\]

Answer 3

log is a function
if you have \(f(x+2)-f(2x-3)=2f(x)\) you cannot divide by \(f\) to get
\[x+2-(2x-3)=2x\]

Answer 4

so I should divide like f(x+2)/(2x-3)?

Answer 5

=2ln(x)?

Answer 6

you can start with \[\ln(x-2)+\ln(2x-3)=2\ln(x) \] and then rewrite it as
\[\ln((x-2)(2x-3))=\ln(x^2)\] using two properties of the logarithm

Answer 7

one is as @kc_kennylau wrote
\[\log(A)+\log(B)=\log(AB)\] that gives you the left hand side
the other is that
\[n\log(x)=\log(x^n)\] which gives you the right hand side

Answer 8

I have ln(2x^2-7x+6)=lnx^2 but I dont know what to progress from there.

Answer 9

Remove the ln from both sides

Answer 10

then, since the log is a one to one function, if \[\ln((x-2)(2x-3))=\ln(x^2)\] then
\[(x-2)(2x-3)=x^2\] and it is algebra from here on in

Answer 11

Oh, after I divide both sides by ln, I subtract the x^2 and factor right?

Answer 12

oh no
you are not dividing by \(\ln\)

Answer 13

the log is a one to one function
that means if \(\ln(A)=\ln(B)\) then you know \(A=B\)
that is not division in any way

Answer 14

Ok.

Answer 15

but yes, subtract \(x^2\) set it equal to zero and solve

Answer 16

so the answer I got is x=6 and 1 , is that correct?

Answer 17

you should get \[x^2-7 x+6 = 0\] or \[(x-6)(x-1)=0\] but now you have to be very careful

Answer 18

the solution to the quadratic is \(x=6\) or \(x=1\) but that is not the solution to your original equation, the one with the log

Answer 19

sometimes there is a negative solution that doesn't fit into the log, but in this case, we have two positive so both would work right?

Answer 20

that is the right idea, but the wrong answer
1 is not a solution, even though it is positive
that is because you have \(\ln(x-2)\) so even though 1 is positive, if you put \(x=1\) you get \(\ln(1-2)=\ln(-1)\) which is impossible

Answer 21

it is not a matter of the number being positive or negative, it is a matter of the actual input being positive or negative
hope this is clear

Answer 22

Okay, I understand now, thank you for taking time to teach me.

Answer 23

yw