How to Solve Logarithmic Equations Quickly and Easily?

[omg precal]

Homework Statement

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

Homework Equations

NONE

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.

 

[rocomath]

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.

you need to show work b4 help is given.

 

[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.

 

[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!

 

[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!

 

[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!

 

[rocomath]

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!

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)

 

[omg precal]

done em all, thanks roco