# logarithmic functions

• Apr 18th 2007, 05:04 PM
confusedagain
logarithmic functions
We have just started this chapter and I am confused on the more complicated problems. The book has omitted the way this is worked out. Could you please help explain the steps to me?

e^3In2-2In5 = 8/25
• Apr 18th 2007, 05:36 PM
ecMathGeek
Quote:

Originally Posted by confusedagain
We have just started this chapter and I am confused on the more complicated problems. The book has omitted the way this is worked out. Could you please help explain the steps to me?

e^3In2-2In5 = 8/25

I assume you need to show that the two sides are equal.

e^(3ln2-2ln5) = e^(ln2^3 - ln5^2) = e^(ln8 - ln25) = e^(ln(8/25) = 8/25

If you need further explanation of certain steps, say so and I'll be happy to help.
• Apr 18th 2007, 05:42 PM
ecMathGeek
Quote:

Originally Posted by ecMathGeek
I assume you need to show that the two sides are equal.

e^(3ln2-2ln5) = e^(ln2^3 - ln5^2) = e^(ln8 - ln25) = e^(ln(8/25) = 8/25

If you need further explanation of certain steps, say so and I'll be happy to help.

I'll go ahead and explain the steps:

First, we have:
e^(3ln2 - 2ln5)

In both terms we have a number being multiplied to the log. We can take each number and make it the exponent of the term inside the log. In other words,
3ln2 = ln2^3 = ln8
2ln5 = ln5^2 = ln25

So we get:
e^(3ln2 - 2ln5) = e^(ln8 - ln25)

Since we have a subtraction of logs, we can combine this into a common log by dividing the terms of the logs. In other words,
ln8 - ln25 = ln(8/25)

So we get:
e^(3ln2 - 2ln5) = e^(ln8 - ln25) = e^(ln(8/25))

Now that we have "e" raised to the ln of some term, we can "cancle" the e and ln since they are considered "inverse opperations." In other words,
e^(lnx) = x
ln(e^x) = x

So we get:
e^(3ln2 - 2ln5) = e^(ln8 - ln25) = e^(ln(8/25)) = 8/25
• Apr 19th 2007, 05:00 PM
confusedagain
I think I understand but...
I think I understand these types of problems now.

So if the problem is e^2ln3 = 3ln2=ln3^2 = 9

What about a problem such as this:
ln e(^3square root e/e^1/3)?
• Apr 19th 2007, 05:12 PM
Jhevon
Quote:

Originally Posted by confusedagain
I think I understand these types of problems now.

So if the problem is e^2ln3 = 3ln2=ln3^2 = 9

What about a problem such as this:
ln e(^3square root e/e^1/3)?

First of all, write out your problems clearly. e^(2ln3) is NOT equal to 3ln2 or ln3^2

you should have written e^(2ln3) = e^(ln(3^2)) = e^(ln9) = 9

keep brackets around your powers if they are expressions, we need to know where they start or stop.

Now we have lne^(3sqrt(e)/e^(1/3))

Now there is a law of logarithms that says: log(x^n) = nlogx

so lne^(3sqrt(e)/e^(1/3)) = (3sqrt(e)/e^(1/3))*lne

there is also a law that says log[a]a = 1, i put the base in [] so i just wrote log to the base a of a = 1
so if the base is the same as the number being logged, the answer is one, i hope you know why (if not ask). Now ln = log[e], so lne = log[e]e = 1 (we don't write log[e]e, we just go straight to 1)

so we have lne^(3sqrt(e)/e^(1/3)) = (3sqrt(e)/e^(1/3))*lne = 3sqrt(e)/e^(1/3)

now sqrt(e) = e^(1/2). A law of exponents says that if we divide a base into itself, we subtract the power of the bottom from the one at the top.

So 3sqrt(e)/e^(1/3) = [3e^(1/2)]/e^(1/3) = 3* e^(1/2) / e^(1/3) = 3 * e^(1/2 – 1/3) = 3e^(1/6) and that's your answer