# logarithm equation

• Apr 15th 2011, 07:23 PM
fran1942
logarithm equation
Hello, I have these two equations but are really stumped as to how to solve them.

ln2 + ln(x+3) = 3

log3 + log(2x-1) = log2 + log(x+1)

I know the concept of multiplying when you have addition, and division when subtracting. In these two cases, I think I would be multiplying the terms as they are added ?

Can someone please get me started on these so I can see how to go about it.

Thanks kindly for any help.
• Apr 15th 2011, 07:27 PM
Prove It
Hint: Simplify everything first using \$\displaystyle \displaystyle \log{(a)} + \log{(b)} = \log{(ab)}\$.
• Apr 15th 2011, 08:48 PM
fran1942
not sure if I am on the right track, but I have factored the two to:

ORIGINAL: log3 + log(2x-1) = log2 + log(x+1)
FACTORED: log3(2x-1) = log2(x+1)

ORIGINAL: ln2 + ln(x+3) = 3
FACTORED: ln2(x+3) = 3

am I correct ?
If so, what is the next step ?

Thanks kindly for any help.
• Apr 15th 2011, 09:08 PM
pickslides
Quote:

Originally Posted by fran1942
FACTORED: log3(2x-1) = log2(x+1)

You can cancel the logs off both sides leaving

3(2x-1) = 2(x+1)

Solve for x.
• Apr 15th 2011, 09:28 PM
fran1942
thanks, I get 5/4 for the first one.

However how can I finish off the second problem:
Do I have to balance the sides first? I dont think I can just drop off the ln on the left.

ORIGINAL: ln2 + ln(x+3) = 3
FACTORED: ln2(x+3) = 3
• Apr 15th 2011, 09:29 PM
Prove It
Quote:

Originally Posted by fran1942
not sure if I am on the right track, but I have factored the two to:

ORIGINAL: log3 + log(2x-1) = log2 + log(x+1)
FACTORED: log3(2x-1) = log2(x+1)

ORIGINAL: ln2 + ln(x+3) = 3
FACTORED: ln2(x+3) = 3

am I correct ?
If so, what is the next step ?

Thanks kindly for any help.

They're both correct. Now you need to exponentiate both sides of both equations.
• Apr 15th 2011, 09:52 PM
fran1942
Quote:

Originally Posted by Prove It
They're both correct. Now you need to exponentiate both sides of both equations.

thanks, would I be correct by doing this:

ORIGINAL: ln2 + ln(x+3) = 3
FACTORED: ln2(x+3) = 3
i) e^3 = 2(x+3)
ii) e^3 = 2x+6
iii) ((e^3) /2)-6 = x
• Apr 15th 2011, 09:56 PM
pickslides
Quote:

Originally Posted by fran1942
FACTORED: ln2(x+3) = 3
i) e^3 = 2(x+3)
ii) e^3 = 2x+6
iii) ((e^3) /2)-6 = x

Nope, you subtract the 6 first then divide by 2.

Do you know why? Its really important you understand this concept!
• Apr 15th 2011, 09:56 PM
Prove It
Not quite...

2x + 6 = e^3
2x = e^3 - 6
x = (e^3 - 6)/2