If x, y, and z are integers, is x even?

[pshu4]

This is from the Number Properties book 5th edition, pg 81, question 1.

If x, y, and z are integers, is x even?

(1) 10^x = (4^y)(5^z)
(2) 3^(x+5) = 27^(y+1)

I am confused as to how the book solves (1). How do you do it?

[ShyT]

Exponent simplification.

Is X even?

10^x= 4^y * 5^z

first off "10^x"="(2^X)(5^X)"

Statement 1 is a FACT. Put it in this form to see if x is even

4^y= 2^2^y= 2^2y

5^z is simplified so the rephrase is

10^x= (2^2y)(5^Z)

for this statement to be true. remember it is true. its a stem.. x=2y=z, 2y is even so x is even z is even. SUFF

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

always simplify

3^x+5= 3^3(y+1)

x+5= 3y+3

subtract 5

x=3y-2

When you become seasoned with your number properties you will immediately notice that this rephrased statement is insufficient because depending on "y" x can be either odd or even. If you can't see it though simply plug in numbers and you'll see x can be either odd or even. INSUFF

A

[jnelson0612]

Thanks! pshu4, please let us know if you need further assistance. :-)

[asingleton08]

I'm sorry but I still don't understand the answer to this question:

Here's what I do "understand" so please correct if I'm wrong.

Statement 1 says 10^x=(4^y)(5^z)

So "10^x"="(2^X)(5^X)"- got that but why does that break down make x an even?

I get that 4^y=2^2y but how does that apply to:

" 5^z is simplified so the rephrase is:
10^x= (2^2y)(5^Z)
for this statement to be true. remember it is true. its a stem.. x=2y=z, 2y is even so x is even z is even. SUFF"


Especially how did you find out x=2y and x=z?

Can someone provide a more detailed explanation?

[tim]

it sounds like you understand that we have (2^2y)(5^z) = (2^x)(5^x). from this (because we're dealing with integers) we can conclude that 2^2y = 2^x, so 2y = x, and since x is two times an integer y, that makes x even. does this help?