Prove that if a^x=b^y=ab^xy then x+y=1

Below.

Step-by-step explanation:

Prove that if

a^x = b^y = (ab)^(xy)  then x + y = 1.

From the second equation:  

y = 1 - x

So we have:

a^x = b^y = b^(1-x)  

Taking logs:

x ln a = (1-x)ln b

x ln a + x ln b = ln b

x = ln b / ( ln a + ln b)

x = ln b / ln ab.

This is true if x + y = 1.

Now consider b^y = (ab)^(xy):

b^(1 - x) = (ab)^(x(1- x)

ln b - x ln b = x(1 - x)ln ab

ln b - x ln b = x ln ab - x^2 ln ab

Now we substitute for x = ln b / ln ab in this expression.

If left side = right side then we have proved this identity and therefore the original one.

Left side =  ln b - (ln b / ln ab) * ln b =  ln b - (ln b)^2 / ln ab.

Right side = (ln b / ln ab) * ln ab - [(ln b)^2 / (ln ab)^2] * ln ab

=    ln b - (ln b)^2 / ln ab.

Left side = right side so this identity is true, and so the original one

( x + y = 1) is true also.

answer: -8.2562234e+13

answer:   b

step-by-step explanation:

x2 = 16

x can not be a negative number because 16 is positive

the only answer would be positive 8