Thread: 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.

Hint: Simplify everything first using .

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.

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.

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
ANSWER: ?

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.

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
ANSWER:
i) e^3 = 2(x+3)
ii) e^3 = 2x+6
iii) ((e^3) /2)-6 = x

Originally Posted by fran1942

FACTORED: ln2(x+3) = 3
ANSWER:
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!

Not quite...

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