Is 2^(y+z)*3^x*5^y*7^z < 90^y*14^z?

The stated inequality can be rewritten as:
2^(y+z) * 3^x * 5^y * 7^z < 2^(y+z) * 3^2y * 5^y * 7^z.

Thus, the real question asked is whether x<2y or not

1) x=1 and y is an integer min. 1. Thus, it is clear that x<2y
SUFFICIENT

2) If y=1 and x=1, then x<2y.
If y=1 and x=10, then x>2y.
NOT SUFFICIENT.

ANSWER IS (A)

Posted from my mobile device

\(2^{(y+z)}*3^x*5^y*7^z < 90^y*14^z....2^y*2^z*3^x*5^y*7^z<9^y*2^y*5^y*2^z*7^z..........(2^y*2^z*5^y*7^z)*3^x<(2^y*2^z*5^y*7^z)*3^{2y}....3^x<3^{2y}\)
So question is -- Is \( 3^x<3^{2y}\)

(1) y and z are positive integers; x = 1
So minimum value of y is 1..
Is \( 3^1<3^{2*1}....3<3^2\)?... No always
Suff

(2) x and z are positive integers; y = 1
Least value of x=1, so Is \( 3^1<3^{2*1}....3<3^2\)?... No
But say x=5, so Is \( 3^5<3^{2*1}....3^5<3^2\)?... Yes
Insuff

A

Question: 2^(y+z)*3^x*5^y*7^z < 90^y*14^z

Rewrite: 2^y*2^z*3^x*5^y*7^z < 2^y*(3^2)^y*5^y*14^z

<=> 2^y*3^x*5^y*14^z < 2^y*3^2y*5^y*14^z
<=> 3^x < 3^2y
<=> x < 2y

(1) y and z are positive integers; x = 1

y > 0 ^ x = 1
Inequality is always sufficient, since smallest value for y = 1 => 1 < 2*1

(2) x and z are positive integers; y = 1

x > 0 ^ y = 1
Not Sufficient since x can also be e.g. 5 => 5 > 2*1

=> A

Is 2^(y+z)*3^x*5^y*7^z < 90^y*14^z?


(1) y and z are positive integers; x = 1

(2) x and z are positive integers; y = 1


We should know the rule n^y*n^z = n^(y+z)

So first of all we have to reformat the equation.
90^y*14^z = (3*3*5*2)^y * (7*2)^z = 3^2y * 5^y * 7^z * 2^(y+z)

Next what I see is that both left and right are equal except the 3 number
2^(y+z) * 3^x * 5^y * 7^z < 2^(y+z) * 3^2y * 5^y * 7^z
which means that we can eliminate all number except 3

So now the question should be changed to
Is 3^x < 3^2y?

Let's look at the question
(1) y and z are positive integers; x = 1
if x = 1
y = positive integers (y = 1,2,3,4,...)
This mean that even y = 1 which is the lowest, 3^x will always less than 3^2y.
x=1, y=1; 3 < 3^2(1) YES
x=1, y=2; 3 < 3^2(2) YES
So (1) is sufficient to answer then (1) is OK.

(2) x and z are positive integers; y = 1
Now we know that y = 1 and x could be 1,2,3,4,...
Let's input the number
x=1, y=1; 3 < 3^2(1) YES
x=2, y=1; 3^2 < 3^2(1) NO it is equal
x=3, y=3; 3^3 < 3^2(1) NO
So (2) is not sufficient to answer. (2) is NOT OK

So i think that the answer should be A

My learning
- when the question is about 2 comparing equation with a lot of xyz, we should reformat the question first before step in to calculate. This usually make everything easier (there are cases when reformat is not working when the answer is obviously full of xyz. So if we see the answer like that, just dont reformat, or else we have to reformat and reformat again just to match the answer choice )
- when we see small equation, input number always be a good strategy

Thanks