SOLUTION: I have a couple of questions that I need help with: 1. Solve for x: ln7 + ln(x-3) = ln(x + 15) 2. Solve for x: 28 = 19 + 12^(x-7) **The (x-7) is an exp

Question 4694: I have a couple of questions that I need help with:

1. Solve for x:
ln7 + ln(x-3) = ln(x + 15)

2. Solve for x:
28 = 19 + 12^(x-7)
**The (x-7) is an exponent for the number 12

3. Solve for x:
e^(3x-4) = 1/e^(x-12)
**The (3x-4) and (x-12) are both exponents

4. Use the properties of logarithms to write 3lnx^2 + ln(2x-4)-ln(3x-6) as a single logarithm.

To whoever is kind enough to help with answering these problems: Thank you and I would appreciate it if you could show your work so I can better understand how to do these types of problems myself. Thanks again!
Found 2 solutions by rapaljer, longjonsilver:
Answer by rapaljer(4667) (Show Source):
You can put this solution on YOUR website!
1. ln7 + ln(x-3) = ln(x + 15)
First get the variables on one side (the right side) by subtracting ln(x-3) from each side.
ln 7 = ln(x+15) - ln(x-3)

Use the property of logs to express the right side as a log of a single quantity:

Because ln A = ln B, you can say that A=B. In the same way,

Multiply both sides of the equation by (x-3):

Check to make sure x= 6 doesn't give you the log of a negative. It does not.

2.
Subtract 19 from each side of the equation:

Take the ln of each side:

By law of logarithms:

By distributive property:

Add 7 ln (12) to each side:

Divide both sides by ln (12)

I'm very sorry, that's all I have time for. Maybe someone else will do the others, or perhaps you can repost them.

R^2 at SCC

Answer by longjonsilver(2297) (Show Source):
You can put this solution on YOUR website!
1.
. Now just raise everything to the power e, to remove the natural logs:

7(x-3) = x+15
7x-21 = x+15
6x = 36
x = 6

2.

. Take logs to the base 10:

x-7 = 0.884228
so x = 7.884228

3.

. Now take natural logs of both sides, to remove the "e":

3x-4 = 12-x
4x = 16
x = 4

--> check these answers in the original questions.

jon.