Question

If 2^x = 3^y = 72^z, find the relation between x, y and z.

Answer 1

Given:

2^x = 3^y = 72^z

Step-wise calculation:

Let the common value be k.

Therefore,

2x = k

⇒ `2 = k^(1/x)`

3y = k

⇒ `3 = k^(1/y)`

72z = k

Note that 72 = 2³ × 3², so:

72z = (2³ × 3²)^z

72z = 2^3z × 3^2z

72z = k

From the expressions for 2 and 3, rewrite k as:

k = 2^x = 3^y

Therefore, k = 2^x = 3^y.

Also, k = 2^3z × 3^2z.

So, 2^x = 2^3z × 3^2z = 3^y.

Comparing the powers of the same base for k, we get the equivalence:

`(k^(1/x))^2 xx (k^(1/y))^3 = k^(1/z)`

This implies `k^(2/x + 3/y) = k^(1/z)`

Since k is positive and arbitrary, equate the exponents:

`(2/x) + (3/y) = 1/z`

The relation between x, y and z is `(2/x) + (3/y) = 1/z`.