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Core 3 Help please

Solve the equation:
ln(x+3) + ln(x-1) = 0

I know you need to take e of them. What I did was,

e^ln(x+3) + e^ln(x-1) = 0

Therefore X+3 + X-1 = 0 hence x = -1 , but the book gives the answer sqrt5 - 1

Thanks for any help in advance! :smile:
Original post by Mitul106
Solve the equation:
ln(x+3) + ln(x-1) = 0

I know you need to take e of them. What I did was,

e^ln(x+3) + e^ln(x-1) = 0

Therefore X+3 + X-1 = 0 hence x = -1 , but the book gives the answer sqrt5 - 1

Thanks for any help in advance! :smile:

This step is wrong, recall that:

log(a)+log(b)=log(ab)log(a)+log(b)=log(ab)

also log(0)=1log(0)=1
Original post by Slowbro93

also log(0)=1log(0)=1


Are you sure?
Reply 3
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Mitul106
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Original post by Slowbro93
This step is wrong, recall that:

log(a)+log(b)=log(ab)log(a)+log(b)=log(ab)

also log(0)=1log(0)=1


Thanks for the quick response I gave it a go but now i end up getting x = -3 or x= 1
Original post by morgan8002
Are you sure?


Silly mistake, ignore :getmecoat:
Reply 5
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MJS107
7
Original post by Mitul106
Solve the equation:
ln(x+3) + ln(x-1) = 0

I know you need to take e of them. What I did was,

e^ln(x+3) + e^ln(x-1) = 0

Therefore X+3 + X-1 = 0 hence x = -1 , but the book gives the answer sqrt5 - 1


Use the addition log rule, where you multiply the brackets to give ln(x+3)(x-1) =0
Expand the brackets
ln(x^2 -2x -3) =0
Now e both sides (e0 is 1)
x^2-2x-3 =1
Now use the quadratic formula on:
x^2-2x-4=0
Which gives
x= sqrt5 -1 and sqrt5+1
(edited 8 years ago)
Original post by Mitul106
Thanks for the quick response I gave it a go but now i end up getting x = -3 or x= 1


Show your working because I've gotten what the book said (I meant to say that e0=1e^0=1
Reply 7
Avatar for Mitul106
Mitul106
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Original post by Slowbro93
Show your working because I've gotten what the book said (I meant to say that e0=1e^0=1


it was me being an idiot i forgot to take e^0 to = 1, sorry to be annoying but is sqrt5 + 1 not possible in the equation because the answer only shows sqrt5 - 1, also I got the answer thanks for the help! :smile:
Original post by Mitul106
it was me being an idiot i forgot to take e^0 to = 1, sorry to be annoying but is sqrt5 + 1 not possible in the equation because the answer only shows sqrt5 - 1, also I got the answer thanks for the help! :smile:


YOu should get:

−1+sqrt5,−1−sqrt5 -1 + sqrt{5}, -1 - sqrt{5} which you can see that only one of them would be a valid solution (can you see why though? :holmes:)
Reply 9
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Mitul106
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Original post by Slowbro93
YOu should get:

−1+sqrt5,−1−sqrt5 -1 + sqrt{5}, -1 - sqrt{5} which you can see that only one of them would be a valid solution (can you see why though? :holmes:)


Ah yes negative sqrt sorry just having a blonde moment on this question for some reason aha Thanks a bunch!
Original post by Mitul106


e^ln(x+3) + e^ln(x-1) = 0



You can do that, in algebra in general if you do something to 1 side you have to do it to the other side. You are not allowed to take e to the power of each term to get rid of the logs, you could take e to the power of the LHS as long as you do the same to the RHS:

e^(ln(x+3) + ln(x-1)) = e^0 = 1

However this does not help us really. Try using the rules of logs to combine ln(x+3) + ln(x-1) into 1 log term. Also remember that a^0 = 1 and therefore Log(1) = 0 .
Original post by Mitul106
Solve the equation:
ln(x+3) + ln(x-1) = 0

I know you need to take e of them. What I did was,

e^ln(x+3) + e^ln(x-1) = 0


On a side note as this has not been addressed really.

What you do to one side of an equation you do to the other. You must do it to the whole of it as one item, not piecemeal.

If you were going to "e" it as this stage, you'd get:

eln(x+3)+ln(x−1)=e0e^{ln(x+3) + ln(x-1)} = e^0

This sort of error has cropped up on here a few times recently; often with squaring
a+b=c a+b=c implying a2+b2=c2a^2+b^2=c^2 which is clearly wrong, and should be (a+b)2=c2(a+b)^2=c^2

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