question1:3^x+2 =5^x-1

question2:ln(x+1)+ln(x-1)=ln8

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- February 21st 2006, 12:04 AMrachaellog
question1:

**3**^x+2 =**5**^x-1

question2:ln(x+1)+ln(x-1)=ln8 - February 21st 2006, 01:01 AMticbolQuote:

Originally Posted by**rachael**

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.

Therefore, x=3 ----------answer.

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. - February 21st 2006, 02:25 AMrachael
thank you