# Thread: quick question on product rule and equality rule for logs

1. ## quick question on product rule and equality rule for logs

so I have an equation:

ln (2x+1) + ln(x-2) -2lnx = 0

so to be able to use product rule I added the 2lnx and cancelled out for equality rule

ln(2x+1)+ln(x-2) = 2 ln x

then ln((2x+1)(x-2)) = 2 ln x, can cancel out the ln so we have

(2x+1)(x-2) = 2x.

Im little confused on what to do next. Should I foil out the left side then add the right? Think im forgetting a step here to fully solve this. Thanks

2. ## Re: quick question on product rule and equality rule for logs

Originally Posted by RYdis
so I have an equation:

ln (2x+1) + ln(x-2) -2lnx = 0

so to be able to use product rule I added the 2lnx and cancelled out for equality rule

ln(2x+1)+ln(x-2) = 2 ln x

then ln((2x+1)(x-2)) = 2 ln x, can cancel out the ln so we have

(2x+1)(x-2) = 2x.

Im little confused on what to do next. Should I foil out the left side then add the right? Think im forgetting a step here to fully solve this. Thanks

Nope. You cannot simply cancel ln(x) while that 2 is infront of it. Your working is good up to $\ln((2x+1)(x-2)) = 2\ln(x)$

Your next step should be to use the log power rule on the RHS

3. ## Re: quick question on product rule and equality rule for logs

Hmm..still having difficulties..guess nit understanding this part on my own

4. ## Re: quick question on product rule and equality rule for logs

Originally Posted by RYdis
so I have an equation:

ln (2x+1) + ln(x-2) -2lnx = 0

so to be able to use product rule I added the 2lnx and cancelled out for equality rule

ln(2x+1)+ln(x-2) = 2 ln x

then ln((2x+1)(x-2)) = 2 ln x, can cancel out the ln so we have

(2x+1)(x-2) = 2x.

Im little confused on what to do next. Should I foil out the left side then add the right? Think im forgetting a step here to fully solve this. Thanks
\displaystyle \begin{align*} \ln{(2x+1)} + \ln{(x - 2)} - 2\ln{(x)} &= 0 \\ \ln{(2x + 1)} + \ln{(x-2)} - \ln{\left(x^2\right)} &= 0 \\ \ln{[(2x+1)(x-2)]} - \ln{\left(x^2\right)} &= 0 \\ \ln{\left[\frac{(2x+1)(x-2)}{x^2}\right]} &= 0 \\ \frac{(2x+1)(x-2)}{x^2} &= e^0 \\ \frac{(2x+1)(x-2)}{x^2} &= 1 \\ (2x+1)(x-2) &= x^2 \end{align*}

Go from here...

5. ## Re: quick question on product rule and equality rule for logs

actually had a slight typo in the problem. But here we go

going off what you had this is what we actually see (x-3 not x-2)

(2x+1)(x-3) = x^2..can we divide out the x's?

[(2x+1)(x-2)]/x^2 = [(2+1)(-3)]/x = 0
(2+1)(-3) = x
x=-9?

everything I know tells me this should be wrong..since x^2 usually means we should have two answers.

However I was unsure on how to factor it..

6. ## Re: quick question on product rule and equality rule for logs

Originally Posted by RYdis
actually had a slight typo in the problem. But here we go

going off what you had this is what we actually see (x-3 not x-2)

(2x+1)(x-3) = x^2..can we divide out the x's?

[(2x+1)(x-2)]/x^2 = [(2+1)(-3)]/x = 0
(2+1)(-3) = x
x=-9?

everything I know tells me this should be wrong..since x^2 usually means we should have two answers.

However I was unsure on how to factor it..
Yes you are wrong in thinking you should divide by x^2. Expand the brackets, move everything to one side, and factorise.

7. ## Re: quick question on product rule and equality rule for logs

yes I see, so we get

2x^2-6x+x-3 = x^2
2x^2-5x-3 = x^2
2x^2-x^2-5x-3 = x^2-5x-3 = 0

this doesn't factor even so x= +- (5+sqrt(37))/2

trying to relearn this on your own is rather difficult lol :/

8. ## Re: quick question on product rule and equality rule for logs

Originally Posted by RYdis
yes I see, so we get

2x^2-6x+x-3 = x^2
2x^2-5x-3 = x^2
2x^2-x^2-5x-3 = x^2-5x-3 = 0

this doesn't factor even so x= +- (5+sqrt(37))/2

trying to relearn this on your own is rather difficult lol :/
I'm sure you mean $\displaystyle x = \frac{5 \pm \sqrt{37}}{2}$... Which is correct

9. ## Re: quick question on product rule and equality rule for logs

yes sir. Thank you very much... complicated lol