Solve for x y log3 x y 3 log5 x y 2 A linear function ha

Solve for (x, y) {log_3 (x + y) = 3 log_5 (x - y) = 2 A linear function has an x-int of square root 3 and a y-int of square root 5. What is the slope of the graph of the function?

Solution

log3(x+y) = 3

log5(x-y) = 2

Use log property: loga(B) = x ----> B = a^x

So. x+y = 3^3

x+y = 27 ----(1)

x-y =5^2 = 25

x- y =25 ----(2)

Add eqautions 1 and 2: we get

2x = 52 ---> x= 26

y = 1

( x, y) -( 26 , 1)

2) X interecpt : ( sqrt3, 0)

Y intercept : ( 0, sqrt5)

Usinjg these two points find the slope:M = ( sqrt5 -0)/( 0 -sqrt3)

= sqrt(5/3)