# log

• Feb 20th 2006, 11:04 PM
rachael
log
question1:3^x+2 = 5 ^x-1
question2:ln(x+1)+ln(x-1)=ln8
• Feb 21st 2006, 12:01 AM
ticbol
Quote:

Originally Posted by rachael
question1:3^x+2 = 5 ^x-1
question2:ln(x+1)+ln(x-1)=ln8

Here is one way.

3^(x+2) = 5^(x-1)
Take the common log, or to the base 10, of both sides,
(x+2)log(3) = (x-1)log(5)
Expand both sides,
x*log(3) +2log(3) = x*log(5) -log(5)
Isolate the x-terms,
x*log(3) -x*log(5) = -log(5) -2log(3)
Too much negatives. Divide both sides by -1,
x*log(5) -x*log(3) = log(5) +2log(3)
x[log(5) -log(3)] = log(5) +log(3^2)
x[log(5/3)] = log(5*9)
x*log(5/3) = log(45)
x = [log(45)]/[log(5/3)] = 7.451980309 -------------answer.

Check,
3^(x+2) = 5^(x-1)
3^(9.451980309) =? 5^(6.451980309)
32,340.0513 =? 32,340.0513
Yes, so, OK.

-------------------------------
ln(x+1) +ln(x-1) = ln(8)
ln[(x+1)(x-1)] = ln(8)
ln(x^2 -1) = ln(8)
x^2 -1 = 8
x^2 = 8 +1
x^2 = 9
x = +,-3 -------***

Check x = 3,
ln(3+1) +ln(3-1) =? ln(8)
ln(4) +ln(2) =? ln(8)
ln(4*2) =? ln(8)
ln(8) =? ln(8)
Yes, so, OK.

Check x = -3,
ln(-3+1) +ln(-3-1) =? ln(8)
ln(-2) +ln(-4) =? .....
Umm, there are no logarithm of negative numbers....
So, no good. Reject x = -3.

Why no logarithm of negative number?
See this example.
If log(-a) = b
Then -a = 10^b.
No amount of b can result to -a.

Another.
ln(-a) = b
-a = e^b
No amount of b can result to -a.

Another.
ln(-a) = -b
-a = e^(-b)
-a = 1/(e^b)
No amount of b can result to -a.
• Feb 21st 2006, 01:25 AM
rachael
thank you