Question

need help.. T.T
Given a+b=2c, evaluate: [2^((a-b)b) * 2^((b-c)(c-a) all over 2^((c+b)(c-b))] ^(1/c)

Answer 1

\[(\frac{2^{(a-b)b} \cdot 2^{(b-c)(c-a)}}{2^{(c+b)(c-b)}})^{^\frac{1}{c}}\]

Answer 2

is that one part write?

Answer 3

yes thats it

Answer 4

are we suppose to write it in terms of just c?

Answer 5

I dont really know.. T.T actually their are choices of the answer.
a. 0
b. 1
c. 2
d. \[2^{2c}\]

Answer 6

Ok let's try something first use law of exponents and then also replace all the c's with (a+b)/2

Answer 7

I mean use the law of exponents on the product part the division part
write just as 2^(some power) inside the parenthesis

Answer 8

I'm telling you to replace all the c's with (a+b)/2 because of the given part which is 2c=a+b

Answer 9

so first do you know what the properties say about simplifying
\[2^m * 2^n\]

Answer 10

\[2^{m+n}\]

Answer 11

ok and what if we had \[\frac{2^m \cdot 2^n}{2^c}\]

Answer 12

\[2^{m+n-c}\]

Answer 13

So you would agree that we could write what you have as:
\[(2^{(a-b)b +(b-c)(c-a)-(c+b)(c-b)})^\frac{1}{c}\]

Answer 14

took top exponents and add and subtracted the bottom exponent

Answer 15

I would multiply everything out and combine any like terms
looking at
\[(a-b)b+(b-c)(c-a)-(c+b)(c-b)\]
Could you do that and let me know what you get ?

Answer 16

\[bc+ac-2c\]

Answer 17

I think you might be missing some terms

Answer 18

so (a-b)b=ab-b^2
(b-c)(c-a)=bc-ab-c^2+ca
(c+b)(c-b)=c^2-b^2

Answer 19

so you need to simplify
\[ab-b^2+bc-ab-c^2+ca-(c^2-b^2)\]
Try simplifying this

Answer 20

oh i see what you did
you left off the square on the c

so you meant to say bc+ac-2c^2

Answer 21

ab-ab-b^2 +b^2 +bc +ca -c^2 -c^2
so only bc+ca-2c^2 remain

Answer 22

ok coolness! :)

Answer 23

Now let't replace all the c's in
\[(2^{bc+ca-2c^2})^{^\frac{1}{c}}\]
with (a+b)/2

Answer 24

so that means we now have:
\[(2^{b \frac{a+b}{2}+\frac{a+b}{2}a-2(\frac{a+b}{2})^2})^{^\frac{2}{a+b}}\]

Answer 25

so we have one law of exponent that says we can do
\[(2^{w+r+s})^v=2^{vw+vr+vs}\]

Answer 26

so that means we have
\[2^{b \frac{a+b}{2} \frac{2}{a+b}+\frac{a+b}{2}a \frac{2}{a+b}-2(\frac{a+b}{2} )^2 \frac{2}{a+b}}\]

Answer 27

you should notice a lot of things cancel

Answer 28

0 is the answer

Answer 29

how did you get that?

Answer 30

is that what the exponent simplified too?

Answer 31

\[b+a-2(\frac{ a+b }{ 2 })\]

Answer 32

b+a-a-b=0
:)

Answer 33

so you would have 2^0 which isn't 0 but ...

hint:
a^0=1 (well whenever we don't have a is 0)

Answer 34

ahhh so it should be 1
thank you very much now i understand..:)

Answer 35

greatness
yes the answer is 1