Question

Given a^(3-x) b^(5x) =a^(x+5) b^(3x) then xlogb/a=?

Answer 1

@satellite73

Answer 2

@amistre64

Answer 3

@traile

Answer 4

i am afraid i do not understand the question

Answer 5

@johnnyappleseed

Answer 6

I got \[x \ln \frac{b}{a}=\ln a\]

Answer 7

a^(3-x) b^(5x) =a^(x+5) b^(3x)
a^(3-x-x-5) = b^(3x-5x)
a^(-2x-2)=b^(-2x)
a^(-2x) a^(-2) = b^(-2x)
a^(-2) = (b/a)^(-2x)
log(a^(-2))=log((b/a)^(-2x))
(-2)loga = -2x log(b/a)
loga =x(log(b /a))

Answer 8

a^(3-x)b^5x= a^(x+5) b^3x
a^(3-x-x-5 )= b^(3x-5x)
a^-2(x+ 1) =b^-2x
then take log on both side
-2(x+1)loga=-2xlogb
then divide it by neg. two
(x+1)loga=xlogb
logb/loga=(x+1)/x
logb/loga= 1 + (1/x) then subtract 1 to the other side
1/x = (logb- loga)/loga
x log(a/b)
Then you will get the answer

Answer 9

@CoCoTsoi plzz help i didnt understand ..

Answer 10

which step? :)

Answer 11

a^(3-x-x-5) = b^(3x-5x)

Answer 12

take a^2 / a^4 as example.
(a)(a)/ ((a)(a)(a)(a)) = 1/(a^2) = a^(-2) = a^(2-4)
Similarly,
a^(3-x) b^(5x) =a^(x+5) b^(3x)
a^(3-x) / a^(x+5) = b^(3x) / b^(5x)
a^(3-x-x-5) = b^(3x-5x)

Answer 13

ok @CoCoTsoi how a^(-2) = (b/a)^(-2x)??

Answer 14

a^(-2x) a^(-2) = b^(-2x)
a^(-2x) a^(-2) / (a^(-2x))= b^(-2x) / (a^(-2x))
a^(-2) = (b/a)^(-2x)