Properties of logarithms

• Mar 11th 2007, 05:18 AM
archistrategos214
Properties of logarithms
Hey guys, we're starting a new lesson on properties of logs tomorrow and I'm wondering if you guys can help my understand it? I wanted to do some advanced reading and I picked 3 random examples from the textbook.

1) ln(x+4/x-2)-2ln(x-2)

2) 1+log312-(1/2)log318-log92

3) 8log(b^2)√x - logb (xy)

BTW, the symbols in smaller font are supposed to mean 'to the base of' so and so. As always, thanks for your help!
• Mar 11th 2007, 08:08 AM
Soroban
Hello, archistrategos214!

If you're just starting, these are awful problems!
I assume you know the basic properties of logarithms.

Quote:

1) .ln[(x+4)/(x-2)] - 2·ln(x - 2)
We have: .ln(x + 4) - ln(x - 2) - 2·ln(x - 2) .= .ln(x + 4) - 3·ln(x-2)

. . = .ln(x + 4) - ln(x - 2)³ .= .ln[(x + 4)/(x - 2)³]

Quote:

2) .1 + log312 - ½·log318 - log92
That "base 9" really messes up the problem!

Assuming you don't know the Base-Change Formula yet,
. . I'll show you a primitive approach.

Let .log
9(2) = p . . . Then: .9^p .= .2 . . (3²)^p .= .2 . . 3^{2p} .= .2

. . Hence: .2p .= .log
3(2) . . p .= .½·log3(2)

Therefore: .log
9(2) .= .½·log3(2)

The problem becomes: .log
3(3) + log3(12) - ½·log3(18) - ½·log3(2)

. . = .log
3(3·12) - ½[log3(18) + log3(2)] .= .log3(36) - ½[log3(18·2)]

. . = .log
3(36) - ½log3(36) .= .½·log3(36) .= .log3(36^½) .= .log3(6)

• Mar 11th 2007, 03:51 PM
archistrategos214
Thanks for the help, Soroban! Yes, we're only starting this lesson today, but hopefully our instructor won't be a sadist and give us questions that are too hard. Again, thanks! :)
• Mar 11th 2007, 09:08 PM
Soroban
Hello again, archistrategos214!

I'll try to explain #3 . . . another ugly problem!

Quote:

3) .8·log(√x) - logb(xy)

Let log
(√x) .= .p . . (b²)^p .= .√x . . b^{2p} .= .√x

Then: .2p .= .log
b(x^½) . . 2p .= .½·logb(x) . . p .= .¼·logb(x)

Hence: .log
(√x) .= .¼·logb(x)

The problem becomes: .8·¼·log
b(x) - logb(xy) . = . 2·logb(x) - logb(xy)

. . = .log
b(x²) - logb(xy) .= .logb(x²/xy) .= .logb(x/y)