Explanation:

Given, 3^x = 5^y = 15^z

We need to show that z(xy) = xy

Step 1:

Expressing 15 in terms of 3 and 5:

15 = 3 x 5

So, 15^z = (3 x 5)^z

= 3^z x 5^z

Step 2:

Substituting the values of 3^x and 5^y in the above expression:

15^z = (3^x)^z x (5^y)^z

= 3^(xz) x 5^(yz)

Step 3:

Now, we can equate the two expressions of 15^z obtained above:

3^(xz) x 5^(yz) = 3^z x 5^z

Dividing both sides by 3^z x 5^z:

(3^x)^z x (5^y)^z / 3^z x 5^z = 1

(3^x / 3^z)^z x (5^y / 5^z)^z = 1

Using the property of exponents, we get:

3^(x-z) x 5^(y-z) = 1

or, 3^(z-x) = 1 / 5^(y-z)

Step 4:

Now, we can substitute the value of 3^(z-x) in the expression 15^z = 3^(xz) x 5^(yz):

15^z = 3^z x 5^z / 5^(y-z)

= 3^z / 5^(z-y)

Multiplying both sides by 3^(xy):

15^z x 3^(xy) = 3^(z+xy) x 5^(xy-z)

Step 5:

Now, we can express 3^(z+xy) in terms of 15^z:

3^(z+xy) = 15^z x 5^(z-xy)

Substituting this in the previous equation:

15^z x 3^(xy) = 15^z x 5^(z-xy) x 5^(xy-z)

Cancelling out 15^z from both sides:

3^(xy) = 5^(2xy-2z)

Taking log on both sides:

log(3^(xy)) = log(5^(2xy-2z))

xy log(3) = (2xy-2z) log(5)

xy log(3) = 2xy log(5) - 2z log(5)

xy = (2xy - 2z) log(5) / log(3)

xy = 2xy log(5) / log(3) - 2z log(5) / log(3)

xy - 2xy log(5) / log(3) = -2z log(5) / log(3)

xy (1-2log(5)/log(3)) = -2z log(5) / log(3)

z = -xy / (2 log(5