x is an integer and x raised to any odd integer is greater

Question

x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity 7^(x-1)-5^x?

(1) z < 25 and w=7^x
(2) x = 4

Bunuel · Answer

x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity 7^(x-1)-5^x?

"x is an integer and x raised to any odd integer is greater than zero" simply means \(x=integer>0\).
The question asks "is w - z greater than 5 times the quantity 7^(x-1)-5^x?".

So, we need to answer whether \(w-z>5(7^{x-1}-5^x)\), given that x is a positive integer.

(1) \(z<25\) and \(w=7^x\)

Is \(7^x-z>5(7^{x-1}-5^x)\)?
Is \(7^x-z>5*7^{x-1}-5^{x+1}\)?
Is \(7^x-5*7^{x-1}+5^{x+1}>z\)?

Since the lowest value of \(x\) is 1, then the lowest value of LHS is when \(x=1\):

\(LHS_{min}=7^x-5*7^{x-1}+5^{x+1}=7-5+25=27\).

So, the lowest value of LHS is 27, which is more than \(z\) as \(z<25\). Therefore, \(7^x-5*7^{x-1}+5^{x+1}>z\) is true. Sufficient.

(2) \(x = 4\). No info about \(w\) and \(z\). Not sufficient.

Answer: A.

tech3 · Answer

Hello, let me try and explain.

paraphrasing the questions gives us, is w-z > 5.7^(x-1) - 5^(x+1). Further, x is a positive integer as an odd power of x is greater than 0.

taking statement 1, you have the question as 7^x-z > 5.7^(x-1) - 5^(x+1)

Now, for all cases where x is a positive integer, 7^x would be greater than 5.7^(x-1). Next, 5^(x+1) would always be >= 25 as the least value of x can be 1.

So, 7^x > 5.7^x-1
and z < 5^(x+1). Reversing this inequality and adding both the inequalities, we get:

7^x - z > 5.7^x-1 - 5^(x+1),

Hence answer A.

Carcass · Answer

the question is not simple but if you think a bit abstract and test number you can arrive to the solution.

The FIRST important thing, breaking the problem, is to understand tha our X is positive. Infact, if a number raised to power of 3 is >0 that means X itself is positive because the odd powr maintain the original sign of the number, so X must be positive.

At this point using the exponent rules we have w - z > 5*7^x-1 - 5^x+1 ----> 7^x - 24 (from stem x<25) > 5*7^x-1 - 5^x+1

testing number 1 ------> 7 -24 > 1 - 25 ---> - 17 > -24 this is TRUE. also if you test some value < 25 positive but also negative the result is the same.

1) sufficient

x = 4 -------> w - z > 5*7^3 - 5^5..........at this point you know nothing about w and z.

2) insufficient

A is the answer.

Kaps07 · Answer

why can't x be zero apart from even ? any odd number raised to the power zero is 1 which is an interger greater than zero. I think x is given as an interger and not only positive integer. if you work with x = 0 , statement 1 is not sufficient ?

Bunuel · Answer

We are told that  integer x raised to to any odd integer is greater than zero :  x^{odd}>0 . Now, if  x=0  then  x^{odd}=0^{odd}=0  (for odd>0), which violates given condition, so  x  cannot be zero.

Hope it's clear.

Kaps07 · Answer

Thanks, I misunderstood the question. I thought it was odd int raised to the power x. It is clear now.

Rajkiranmareedu · Answer

Please explain why the lowest value of X is 1.

Bunuel · Answer

See the highlighted part:  x is an integer and x raised to any odd integer is greater than zero  means that x is a positive integer, thus its lowest possible value is 1.

Hope it's clear.

KarishmaB · Answer

Responding to a pm:

What is the meaning of: x is an integer and x raised to any odd integer is greater than zero?
It is just a convoluted way of saying 'x is a positive integer'. Since x raised to any odd power is positive, x must be positive. x can take values 1, 2, 3, 4 and so on.

Question: Is  w - z > 5*7^{x-1}-5^{x+1}

Using statement 1: Is  7^x - 25 > 5*7^{x-1} - 5^{x+1}

Even though we are given that z is less than 25, let's assume it to be 25. In case we can prove that left hand side is greater than right hand side for z = 25, we can prove that left hand side will remain greater than right hand side for any other value of z. If z has a smaller value, say 9, left had side will become even larger (since a smaller number will be subtracted) than the right hand side.

Now note that  7^x  will be greater than  5*7^{x - 1}  since  7^x  can be written as  7*7^{x - 1}  (x is a positive integer).
Also note that minimum value of x is 1 so  5^{x + 1}  will always be equal to or larger than 25 (x is a positive integer).

So comparing the left and right hand side expressions,  7^x  is greater than  5*7^{x - 1}  and 25 is less than or equal to  5^{x+1} . So left hand side must be greater than the right hand side in all cases when x is a positive integer.

Answer (A)

BrentGMATPrepNow · Answer

Given : x is an integer and x raised to any odd integer is greater than zero KEY CONCEPT: A number raised to an ODD power retains its sign. That is, POSITIVE^ODD = POSITIVE and NEGATIVE^ODD = NEGATIVE The given info tells us that x^ ODD = POSITIVE So, it must be the case that x is a POSITIVE integer Target question : Is w - z > 5 Let's tidy the target question by expanding the right side of the inequality to get.... REPHRASED target question : Is w - z > 5 - 5^(x+1)? Statement 1 : z < 25 and w = 7^x Take the REPHRASED target question and replace w with 7^x to get: Is 7^x - z > 5 - 5^(x+1)? IMPORTANT: Notice that, on the left side of the inequality, we're subtracting z We're told that z < 25 So, we can help MINIMIZE the value of the left side by making z as big as possible So, let's make z = 25 (aside: we could make z equal something really close to 25, like 24.99999999999, but let's make things easy on ourselves and make z EQUAL 25. We get: Is 7^x - 25 > 5 - 5^(x+1)? Now let's get like terms on the same side of the inequality. First subtract 5 from both sides to get: Is 7^x - 5 - 25 > -5^(x+1)? Then add 25 to both sides to get: Is 7^x - 5 > 25 - 5^(x+1)? Factor 7^(x-1) from left side to get: Is 7^(x-1) > 25 - 5^(x+1)? Simplify left side to get: Is 7^(x-1) > 25 - 5^(x+1)? Now recognize that 7^(x-1) will be POSITIVE for all values of x. So, 7^(x-1) must be POSITIVE Now recognize that since x is a POSITIVE integer , 5^(x+1) must be greater than or equal to 25 (since x = 1, is the smallest possible value of x) So, 25 - 5^(x+1) must be less than or equal to zero. So, our question becomes Is some POSITIVE number > some number that's less than or equal to zero? The answer to this REPHRASED target question is a definitive YES Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT Statement 2 : x = 4 Since we have no information about z or w, we cannot answer the REPHRASED target question with certainty. So, statement 2 is NOT SUFFICIENT Answer: A Cheers, Brent

Kimberly77 · Answer

Thanks Brent  BrentGMATPrepNow  for detaailed explanation, To clarify this part of 5^x > 5^(x+1), Is it 5 times the quantity 5(7^(x-1)-5^x)? Therefore we've 1 extra 5 here when equation expanded ? Thanks Brent

Target question: Is w - z > 5 
Let's tidy the target question by expanding the right side of the inequality to get....
REPHRASED target question: Is w - z > 5  - 5^(x+1)?

BrentGMATPrepNow · Answer

Thanks Brent  BrentGMATPrepNow  for detaailed explanation, To clarify this part of 5^x > 5^(x+1), Is it 5 times the quantity 5(7^(x-1)-5^x)? Therefore we've 1 extra 5 here when equation expanded ? Thanks Brent

Target question: Is w - z > 5 
Let's tidy the target question by expanding the right side of the inequality to get....
REPHRASED target question: Is w - z > 5  - 5^(x+1)?

Key property: a(b + c) = ab + ac
In the expression  , we have two terms: 7^(x-1) and 5^x. 
You can consider these terms b and c (as in the property a(b + c) = ab + ac)

In the expression 5 , I am multiplying each of the two terms by 5. 
When we do so we get: 5  - 5(5^x)
If we focus on the second part, 5(5^x), we get: 
5(5^x) = (5^1)(5^x) = 5^(x+1)

Does that help?

Kimberly77 · Answer

Great explanation always !! Thanks  BrentGMATPrepNow

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x is an integer and x raised to any odd integer is greater

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