Question

3^x = 5y = (75)^z , show that z = xy/2x+y

Can anyone please help me with this

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Answer 1

Step-by-step explanation:

Let 3^x = 5y = (75)^z=k

So, 3 = k^(1/x); 5 = k^(1/y) and 75 = k^(1/z)

75 = 3 x 25 = 3 x 5²

Substituting the values from step above,

k^(1/z) = {k^(1/x)}*{k^(2/y)} = k^(1/x + 2/y) [By the law of exponents]

Equating the powers from both sides,

1/z = 1/x + 2/y = (y + 2x)/xy

Taking reciprocal, z = (x*y)/(2x + y) [Proved]

Answer 2

Step-by-step explanation:

Given, 3^x = 5y = (75)^z

Let 3^x = 5y = (75)^z = a

=> 3^x = a => 3 = a^1/x ------------ (1)

5y = a => 5 = a^1/y

=> (5)² = a^²/y => 25 = a^2/y ------------ (2)

(75)^z = a => 75 = a^1/z ------------- (3)

from the above equations (1) & (2)

( 3 ) ( 25 ) = ( a^1/x ) ( a^2/y )

=> 75 = a^(1/x + 2/y) [ using a^m x a^n = a^m+n ]

=> a^1/z = a^(2x+y/xy)

Here, bases are same => powers also equal

=> 1/z = 2x+y/xy

=> z = xy / 2x+y