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

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.

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.

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.

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 ?

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[7^(x-1) - 5^x?]
Let's tidy the target question by expanding the right side of the inequality to get....
REPHRASED target question: Is w - z > 5[7^(x-1)] - 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[7^(x-1)] - 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[7^(x-1)] - 5^(x+1)?

Now let's get like terms on the same side of the inequality.
First subtract 5[7^(x-1)] from both sides to get: Is 7^x - 5[7^(x-1)] - 25 > -5^(x+1)?
Then add 25 to both sides to get: Is 7^x - 5[7^(x-1)] > 25 - 5^(x+1)?
Factor 7^(x-1) from left side to get: Is 7^(x-1)[7 - 5] > 25 - 5^(x+1)?
Simplify left side to get: Is 7^(x-1)[2] > 25 - 5^(x+1)?

Now recognize that 7^(x-1) will be POSITIVE for all values of x.
So, 7^(x-1)[2] 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

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

Please explain why the lowest value of X is 1.

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.

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[7^(x-1) - 5^x?]
Let's tidy the target question by expanding the right side of the inequality to get....
REPHRASED target question: Is w - z > 5[7^(x-1)] - 5^(x+1)?

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[7^(x-1) - 5^x?]
Let's tidy the target question by expanding the right side of the inequality to get....
REPHRASED target question: Is w - z > 5[7^(x-1)] - 5^(x+1)?[/quote]

Key property: a(b + c) = ab + ac
In the expression [7^(x-1) - 5^x], 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[7^(x-1) - 5^x], I am multiplying each of the two terms by 5.
When we do so we get: 5[7^(x-1)] - 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?

Great explanation always !! Thanks BrentGMATPrepNow

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