Question

if x and y are real numbers such that 7^x -16y=0 and 4^x-49y=0 , then the value of y-x is

Answer 1

Given, 7^x - 16y = 0 => 7^x = 16y .....(1)
4^x - 49y = 0 => 4^x = 49y ..........(2)

dividing equation (1) by (1) ,

7^x/4^x = 16y/49y = (4/7)^2

(7/4)^x = (7/4)^-2

x = -2

put x = -2 in equation (1),
7^-2 - 16y = 0

y = 7^-2/16 = 1/(49 × 16)

so, y - x = 1/(49 × 16) - (-2)

= 1/(49 × 16) + 2

= (1 + 49 × 32)/(49 × 16)

= 2.00127551

Answer 2

HEY MATE !


HERE'S YOUR ANSWER


ANSWER :-

2.00127551


EXPLAINATION :-


7^x - 16y = 0

=> 7^x = 16y

4^x - 49y = 0

=> 4^x = 49y

Divide these equation ,

7 ^ x / 4 ^ x = 16 y / 49y = ( 4 / 7 ) ^ 2

( 7 / 4 ) ^ x = ( 7 / 4 ) ^ - 2

x = - 2

Put x = -2 in this

7 ^ - 2 - 16 y = 0

y = 7 ^ - 2 / 16 = 1 / ( 49 × 16 )

So,

=> y - x = 1 / ( 49 × 16 ) - ( - 2 )

=> 1 / ( 49 × 16 ) + 2

=> ( 1 + 49 × 32 ) / ( 49 × 16 )

=> 2.00127551