# logarithmic problems

• February 1st 2007, 10:24 AM
kcsteven
logarithmic problems
I have been working on these problems and I must have a logic problem, could you put me in the right direction? Here is the problem and what I have done:
If A=2, B=3, C=5 I am supposed to evaluate the following:
ln(A^3/B^-3C^3)
ln(A^3)-ln(B^-2)-ln(C^3)
3lnA + 2lnB - 3lnC
3ln(2) + 2ln(3) - 3ln(5) = -5516476183
The back of my calc book tells me the answer is incorrect, could you let me know where I went wrong.
Thank you!!!!!!!!!!!!!!!!!
Keith Stevens:)
• February 1st 2007, 10:42 AM
Soroban
Hello, Keith!

Did you leave out part of the problem?

Quote:

If $A=2,\:B=3,\: C=5$, evaluate: . $\ln\left(\frac{a^3}{b^{-3}c^3}\right)$

There is a difference between $A$ and $a$.

Is $A$ somehow related to $a$ ?
If not, it's a truly stooopid problem . . .

Here's another: .If $P = 2$ and $Q = 7$, evaluate $x + y$

• February 1st 2007, 11:30 AM
topsquark
Quote:

Originally Posted by kcsteven
3ln(2) + 2ln(3) - 3ln(5) = -5516476183

What??? You DID mean: -0.5516476183, yes?

Quote:

Originally Posted by kcsteven
If A=2, B=3, C=5 I am supposed to evaluate the following:
ln(A^3/B^-3C^3)

Your problem is a slight typo in the second term:
$ln \left ( \frac{A^3}{B^{-3}C^3} \right )$

$3ln(A) - (-3)ln(B) - 3ln(C) = 3ln(2) + 3ln(3) - 3ln(5) =\approx 0.546965$

-Dan