# Homework Help: Some Annoying Logarithms

1. Aug 22, 2007

### omg precal

1. The problem statement, all variables and given/known data

3^(x - 2) = 30

(x - 2)^3 = 30

log (base 4) (x + 3) + log (base 4) (x - 3) = 2

the square root of: (4^(x + 3) / 16^x) = 32

log (base 9) x = 1.5

2. Relevant equations

NONE

3. The attempt at a solution

I haven't done this since last year. Someone refresh my memory. Please show me how to do it and fast fast fast!!! Thanks in advance.

2. Aug 22, 2007

### rocomath

you need to show work b4 help is given.

3. Aug 22, 2007

### omg precal

i struggled but finally got every one but this one:

log (base 4) (x + 3) + log (base 4) (x - 3) = 2

just get me started please. thanks.

4. Aug 22, 2007

### omg precal

i know there is a rule that if they are the same base logs, u can do something to them but idk really!

5. Aug 22, 2007

### rocomath

the way logs are read ... log(base)y=x

the base must be raised to some power x to get y

show an attempt at all of your questions and i will work them all out!

6. Aug 22, 2007

### omg precal

ok i think i vaguely remember that if u add logs with the same base, you can multiply what is in the log. that woiuld make it:

log base 4 of x^2 - 9 = 2

meaning 16 = x^2 - 9

meaning x^2 = 25

meaning x = 5

hooray, thanks!

7. Aug 22, 2007

### rocomath

excellent :) the first 2 are ez since now you've solved this one ... the square root one, i'll work

square both sides

4^(x+3)/4^(2x) = 32^2

4^(x+3-2x) = 32^2 (simplify your exponent, take the log of both sides, use one of the rules of the power rule, bring down the exponent)

x = 3 - log(32^2)/log(4)

Last edited: Aug 22, 2007
8. Aug 22, 2007

### omg precal

done em all, thx roco