Hi rastis

1) 10^x = (4^y)(5^z)
=>(2*5)^x = 2^(2y) * 5^z
=>2^x * 5^x = 2^(2y) * 5^z

Comparing the powers of 2 we get x = 2y

Therefore x is even. Suff

2) 3^x+5 = 27^y+1

I am assuming it means 3^(x+5) = 27^ (y+1)
=> 3^(x+5) = 3^3(y+1)

Comparing powers of 3

x+5 = 3y + 3
=>x= 3y-2

For y = 0 , x= -2 .... Even
For y = 1, x= 1...... Odd

THerefore Insuff

ANS A

hope it helps

Hi rastis

1) 10^x = (4^y)(5^z)
=>(2*5)^x = 2^(2y) * 5^z
=>2^x * 5^x = 2^(2y) * 5^z

Comparing the powers of 2 we get x = 2y

Therefore x is even. Suff

2) 3^x+5 = 27^y+1

I am assuming it means 3^(x+5) = 27^ (y+1)
=> 3^(x+5) = 3^3(y+1)

Comparing powers of 3

x+5 = 3y + 3
=>x= 3y-2

For y = 0 , x= -2 .... Even
For y = 1, x= 1...... Odd

THerefore Insuff

ANS A

hope it helps

What happened to the 5^z? What do you mean about comparing 2? you lost me

Hi rastis

Let me start from the basic. In a question like this,

1. You need to reduce the terms on both side of the equation in terms of prime factors.
That is what I have done here.
10^x = (4^y)(5^z) =>(2*5)^x = 2^(2y) * 5^z

2. You need to compare the powers/ exponent of the common bases.
In this case, we have both sides in powers of 2 and 5

Comparing powers of 2 , we get x=2y
Comparing powers of 5 , we get x=z

Now, the question asks us whether x is even or not.

From x= z, we can't say as we don't know the value of z
But, from x=2y , x will always be even irrespective of the value of y

Hope it's clear now

A LOT clearer. Thanks!!!!!!!!