Simplify the expression attached and write it as a single logarithm.. ***still confused on this topic... pls help? :)
answer choices A,B,C,D from top to bottom :)
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.
im confused.. so how can i apply that with my expression?
Confused with what? \(-3\log(x+4) = \log\left[\left(x+4\right)^{-3}\right]\) Correct?
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 :(
Your problem statement begins \(-3\log(x+4)\), doesn't it? Or am I looking at some other picture?
yup iy does :) so i get log[(x+4)^−3] from that part?
Very good. Now the (x-7) part?
2log(x-7) = log(x-7)^2 ??
Good, now the x-2...
5log(x-2)=log(x-2)^5 ??
One more. You're on a roll!
lol thanks haha :P but i don't get this one... it looks different.. :/ -logx^2...
but heres my take on it.. log(x^2)^-1 ??
Take the negative (-1) inside, just like the other coefficients ==> exponents.
was what i got right?? or no?
or is it log(x)^-2 ?
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)
sorry my computer crashed... one sec :/
log[(x+4)^−3]+ log(x-2)^5+log(x-7)^2 + log(x)^-2=answer B right? :)
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 ]
kk I'm seeing that :) but how can i simplify it even further?
(x+4)^−3 is really 1/[ (x+4)^3 ]
same idea applies to (x)^-2
ok so i get 1/x^-2 ?
yep
so put that all together
so i get answer C right?? :)
sorry i meant answer D :) is it answer D then? :)
@jim_thompson5910 :)
yes it is D
kk great!!! thx :)
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