iheartfood
Simplify the expression attached and write it as a single logarithm..
***still confused on this topic... pls help? :)
Delete
Share
This Question is Closed
iheartfood
Best Response
You've already chosen the best response.
1
iheartfood
Best Response
You've already chosen the best response.
1
answer choices A,B,C,D from top to bottom :)
tkhunny
Best Response
You've already chosen the best response.
0
Move all the exponents up:
\((x+4)^{-3}\)
\((x-7)^{2}\)
\((x-2)^{5}\)
\((x^{2})^{-1}\)
Notice how I deliberately left everything outside as ADDITION. This allows us just to bring everything inside without concern for numerator or denominator. We can figure all that out later.
iheartfood
Best Response
You've already chosen the best response.
1
im confused.. so how can i apply that with my expression?
tkhunny
Best Response
You've already chosen the best response.
0
Confused with what? \(-3\log(x+4) = \log\left[\left(x+4\right)^{-3}\right]\) Correct?
iheartfood
Best Response
You've already chosen the best response.
1
im not quite sure what we are working on now.. I'm like uber confused.. are we doing my problem? or an example? :/ sorry I'm a bit confused :(
tkhunny
Best Response
You've already chosen the best response.
0
Your problem statement begins \(-3\log(x+4)\), doesn't it? Or am I looking at some other picture?
iheartfood
Best Response
You've already chosen the best response.
1
yup iy does :) so i get log[(x+4)^−3] from that part?
tkhunny
Best Response
You've already chosen the best response.
0
Very good. Now the (x-7) part?
iheartfood
Best Response
You've already chosen the best response.
1
2log(x-7) = log(x-7)^2 ??
tkhunny
Best Response
You've already chosen the best response.
0
Good, now the x-2...
iheartfood
Best Response
You've already chosen the best response.
1
5log(x-2)=log(x-2)^5 ??
tkhunny
Best Response
You've already chosen the best response.
0
One more. You're on a roll!
iheartfood
Best Response
You've already chosen the best response.
1
lol thanks haha :P but i don't get this one... it looks different.. :/
-logx^2...
iheartfood
Best Response
You've already chosen the best response.
1
but heres my take on it..
log(x^2)^-1 ??
tkhunny
Best Response
You've already chosen the best response.
0
Take the negative (-1) inside, just like the other coefficients ==> exponents.
iheartfood
Best Response
You've already chosen the best response.
1
was what i got right?? or no?
iheartfood
Best Response
You've already chosen the best response.
1
or is it log(x)^-2 ?
tkhunny
Best Response
You've already chosen the best response.
0
There it is. They are now all connected by addition. Use this guy and bring them all together. log(a) + log(b) = log(a*b)
iheartfood
Best Response
You've already chosen the best response.
1
sorry my computer crashed... one sec :/
iheartfood
Best Response
You've already chosen the best response.
1
log[(x+4)^−3]+ log(x-2)^5+log(x-7)^2 + log(x)^-2=answer B right? :)
jim_thompson5910
Best Response
You've already chosen the best response.
1
use the rule described above to go from
log[(x+4)^−3]+ log(x-2)^5+log(x-7)^2 + log(x)^-2
to
log[ (x+4)^−3*(x-2)^5*(x-7)^2*(x)^-2 ]
iheartfood
Best Response
You've already chosen the best response.
1
kk I'm seeing that :) but how can i simplify it even further?
jim_thompson5910
Best Response
You've already chosen the best response.
1
(x+4)^−3 is really 1/[ (x+4)^3 ]
jim_thompson5910
Best Response
You've already chosen the best response.
1
same idea applies to (x)^-2
iheartfood
Best Response
You've already chosen the best response.
1
ok so i get 1/x^-2 ?
jim_thompson5910
Best Response
You've already chosen the best response.
1
so put that all together
iheartfood
Best Response
You've already chosen the best response.
1
so i get answer C right?? :)
iheartfood
Best Response
You've already chosen the best response.
1
sorry i meant answer D :) is it answer D then? :)
iheartfood
Best Response
You've already chosen the best response.
1
@jim_thompson5910 :)
jim_thompson5910
Best Response
You've already chosen the best response.
1
yes it is D
iheartfood
Best Response
You've already chosen the best response.
1
kk great!!! thx :)