Write an expanded logarithm into a single logarithm (all rule are reversible)? Help :(?

Question

Expanded log shows as follows:
2log_b (12) - 4 + log_b (1)

by paper says HINT:
Write -4 lob_b (b) in place of -4

so...

log_b (12)^2 - log_b (b) + log (b) 1

according to log rules...
log_b (b) = 1...or in this case, -1...right? Or am I wrong?

according to log rules...
log (b) 1 = 0...right? Or am I wrong?

so...

log_b (12)^2 - log_b (b) + log (b) 1 goes to...

log_b (12)^2 - 1 + 0 [question...why should I listen to the hint? How exacly is that even true? Explain please :()

so..

(12^2)
lob_b -----------
-1

Am I doing this write? please help me...I have a final for pre-cal tomorrow... :(

answers yahoo messed the format of my answer...

log_b ( [12^2] / -1 )

Answer 1

2log_b (12) - 4 + log_b (1)
-4 = log_b[b^(-4)]
2log_b (12) - 4 + log_b (1) = log_b{(12^2)[b^(-4)](1)}
= log_b{144/b^4}

Answer 2

2log_b(12)-4+log_b(1)=
2log_b(12)-4log_b(b)+log_b(1)=
log_b(12^2)-log_b(b^4)+log_b(1)=
log_b[(12^2)(1)/(b^4)]=
log_b(144/b^4)

Note:
log_b(1)=0
log_b(b)=1.
You have to express all the terms of the sum as log with the same base

Answer 3

All rule are NOT reversibles
log a +logb = log(ab) but log(a*b) is nor always log a+log b
as a and b can be negative so a*b is positive,has a log
but loga and logb don´t exist
If you know that a*b is positive but don´t know if a and b are
log(ab) = lod IaI+logIbI
In the same way for the quotient
In the same way log x^2 is not 2 logx unless x >0
It is always 2 log IxI
Be very careful