If n is a positive integer, is the value of b - a at least twice the

For statement (1), I understand how you arrived at \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\) but how does this equation tell you whether b-a is at least twice the value of 3^n-2^n?

For example, if i sub in n=1 into the equation, it is true that b-a is at least twice the value (b-a= 5 and 3^n-2^n = 1 then 1*2). However, if i sub in n=2, the equation no longer is true as b-a then becomes 19 and the right side of the equation then becomes 10, which is not at least twice the value.

What am i missing here?

For the first statement after some manipulation the question becomes: is \(3^{n}\geq{0}\)? Irrespective of the actual value of n, 3^n will always be greater than zero, thus the answer to the question is YES.

The Question asks: b - a > 2*(3^n - 2^n)

The answer should be a definitive yes/no

Statement 1:

a= 2^(n+1) and b= 3^(n+1)

Lets take the smallest possible value of a Positive Integer i.e 1 and put it in for "n"

3^(1+1) - 2^ (1+1) > 2*(3^1 - 2^1)
3^2 - 2^2 > 2* (3 - 2)
9 - 4 > 2 *1
5>2 (a definitive answer)

Lets test another number (just to be on the Safe Side), lets test n=2

3^(2+1) - 2^ (2+1) > 2*(3^2 - 2^2)
3^3 - 2^3 > 2* (9 - 4)
27 - 8 > 2* 5
19> 10 (a definitive answer)

Thus, Sufficient.

Statement 2:

n=3

Let's Put it in the in-equality b - a > 2*(3^n - 2^n)

b - a> 2*(3^3 - 2^3)
b - a> 2*(27 - 8)
b - a> 2*(19)
b - a> 38

If, b=100 and a=10, than definitive answer
If, b= 2 and a= 10, than definitive answer
But since it doesn't provide any value for either "a" or "b"

Thus, Not Sufficient

Therefore the answer is

Can someone please show me the simplification as to how the equation becomes whether 3^n > 0?

Thanks a lot Bunuel. The explanation couldn't have been more lucid.

Hi Bunuel,
In the end instead of subtracting \(2*3^n\) from both sides, can we cancel \(3^n\) from both sides (as it's always positive) and reach \(3>2\) making the statement sufficient? Is that also a correct approach?
Thanks.

Yes, this also would be a correct way of solving.

Target question: Is \(b - a \geq 3^n - 2^n\)?

Statement 1: \(a= 2^{n+1}\) and \(b= 3^{n+1}\)

Substitute values into target question to get: Is \(b= 3^{n+1}- 2^{n+1} \geq 3^n - 2^n\)?

Rearrange terms: Is \(3^{n+1}- 3^n \geq 2^{n+1}-2^n \)?

Factor both sides: Is \(3^n(3- 1) \geq 2^n(2- 1)\)?

Simplify: Is \(3^n(2) \geq 2^n\)?

Divide both sides of the inequality by 2 to get: Is \(3^n \geq 2^{n-1}\)?

At this point, it's clear that the left side is greater than the right side (since the left side of the inequality is the product of n 3's, while the right side of the inequality is the product of n-1 2's)

As such we can be certain that \(3^n \geq 2^{n-1}\)
Statement 1 is SUFFICIENT


Statement 2: n = 3
Since we have no information about the variables a and b, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

Hey - can someone explain the 'cancel' step? I am lost on how the cancelling is done here, thx.

Notice that we have like terms, \(-2*2^{n}\), in both sides of \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}}\), so we can cancel them. Add \(2*2^{n}\) to both sides to get \(3*3^{n}\geq{2*3^n}\).

Bunuel

I'm probably missing something really obvious here, but if we plug in n = 2 for statement 1:

\(3^3 - 2^3 = 19\)

\(2 (3^2 - 2^2) = 10\)

19 is not twice the value of 10. Don't we get yes and no answers?

19 is at least twice the value of 5 (3^2 - 2^2).

Is the value of b - a at least twice the value of 3^n - 2^n?

If n = 2, then:

b - a = 19
3^n - 2^n = 5. Twice the value of 5 is 10.

19 > 10.

Ah, yes. Silly mistake indeed.

Thank you Bunuel.

Hi Bunuel,

Can you please explain the step highlighted in red

\(3^{n+1}=3*3^n\);

\(2^{n+1}=2*2^n\);

\(2*(3^n - 2^n)=2*3^n - 2*2^n\).

Or you meant something else?

I was trying to understand how LHS = RHS in this particular step \(3^{n+1}=3*3^n\), which was quite confusing for me.

we can basically remove +1 from the exponent, and simply multiply the coefficient with the same value.

for example
\(7^{n+1}=7*7^n\)

or

\(7^{n+2}= 7*7*7^n\)


or

\(7^{n+3}= 49*7*7^n\)

is my understanding correct?


and thanks for the previous explanation!

Yes, generally \(a^n*a^m=a^{n+m}\).

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