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Re: x<0 [#permalink]
1
Asmakan wrote:
I tried
x=-1, -.5,-10

The answer was B.
I read the explanation but couldn't understand it. Because they create a relation between 5^x and 4^x. Which really I couldn't understand


You forgot that x can be fractions. Unless you can mathematically prove the relationship, the answer is D.
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Re: x<0 [#permalink]
Just edit the question and put the range. For the 5 it is raised to (x+1) power
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Re: x<0 [#permalink]
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OMG :roll:

Edit the question above. Please Sir, provide always the question ID when it comes from our test.

So I can edit in case.

Thank you so much
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Re: x<0 [#permalink]
Okay.
Now may someone explain it for me ?
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Re: x<0 [#permalink]
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Sir, here on the board we have rules that must be followed. The rules are there to help the students and create order and progresso.

I know you are probably pressured by the coming exam, but this is not the way we deal with. Considering the daily efforts we made to help you

Now

QA is \(4^x\)

QB is \(5^5 \times 5\)

For logic there is no way A could be > B or the two are equal.

Even picking numbers.

x must be negative or an integer or a fraction

\(x =-1\)

\(4^{-1} = \frac{1}{4}\)

\(5^{-1+1} = 5^0 = 1\)

\(B > A\)

\(x=-\frac{1}{2}\)

\(4^{-\frac{1}{2}}= \sqrt{4^-1} = \sqrt{\frac{1}{4}} = 0.5\)

\(5^{- \frac{1}{2} + 1 }= 5^{\frac{1}{2}} = 5 = 2.2\)

\(B > A\)
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x<0 [#permalink]
4
Asmakan wrote:
\(x<0\)

Quantity A
Quantity B
\((4)^x\)
\((5^{x+1})\)




Useful property #1: \(\frac{x^k}{y^k}=(\frac{x}{y})^k\)

Useful property #2: \((\frac{x}{y})^{-k}=(\frac{y}{x})^{k}\)

Given:
Quantity A: \((4)^x\)

Quantity B: \(5^{x+1}\)


Rewrite Quantity B as follow:
Quantity A: \(4^x\)

Quantity B: \((5^x)(5^1)\)


Divide both quantities by \(5^x\) to get:
Quantity A: \(\frac{4^x}{5^x}\)

Quantity B: \(5^1\)


Simplify both quantities to get:
Quantity A: \((\frac{4}{5})^x\)

Quantity B: \(5\)


Since we're told that x < 0, \((\frac{4}{5})^x\) can become infinitely large. Here's what I mean.

If x = -1, then \((\frac{4}{5})^x=(\frac{4}{5})^{-1}=(\frac{5}{4})^{1}=\frac{5}{4}=1.25\)

If x = -2, then \((\frac{4}{5})^x=(\frac{4}{5})^{-2}=(\frac{5}{4})^{2}=\frac{25}{16}≈1.56\)

If x = -3, then \((\frac{4}{5})^x=(\frac{4}{5})^{-3}=(\frac{5}{4})^{3}=\frac{125}{64}≈1.95\)

As we can see, \((\frac{4}{5})^x\) can become super big.
For example,
If x = -20, then \((\frac{4}{5})^x=(\frac{4}{5})^{-20}=(\frac{5}{4})^{20}≈86.7\)

Since Quantity A can range from 1.25 to infinity, the correct answer is D.

Cheers,
Brent




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Re: x<0 [#permalink]
Expert Reply
Why Sir picking numbers above as my strategy has a flaw ?'

Where is the catch ??
Also, I said in a previous post D because the user did not post the constraint x < 0


Thank you Sir
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Re: x<0 [#permalink]
Carcass wrote:
Sir, here on the board we have rules that must be followed. The rules are there to help the students and create order and progresso.

I know you are probably pressured by the coming exam, but this is not the way we deal with. Considering the daily efforts we made to help you

Now

QA is \(4^x\)

QB is \(5^5 \times 5\)

For logic there is no way A could be > B or the two are equal.

Even picking numbers.

x must be negative or an integer or a fraction

\(x =-1\)

\(4^{-1} = \frac{1}{4}\)

\(5^{-1+1} = 5^0 = 1\)

\(B > A\)

\(x=-\frac{1}{2}\)

\(4^{-\frac{1}{2}}= \sqrt{4^-1} = \sqrt{\frac{1}{4}} = 0.5\)

\(5^{- \frac{1}{2} + 1 }= 5^{\frac{1}{2}} = 5 = 2.2\)

\(B > A\)

I thought you got it because you fix it.

Of course, I will put it always
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Re: x<0 [#permalink]
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No.

The question has the OA in the test as D. In fact, the answer is D.

However, you posted as B. And I picked it also wrong. Regardless what was the OA

All this mess is because of the question was posted wrongly at the beginning. Wrong the OA, without the constraint x < 0.

In doing so, we create more damage than benefit.

regards
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Re: x<0 [#permalink]
Ok

Posted from my mobile device Image
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Re: x<0 [#permalink]
2
Carcass wrote:
Why Sir picking numbers above as my strategy has a flaw ?'

Where is the catch ??
Also, I said in a previous post D because the user did not post the constraint x < 0


Thank you Sir


Your solution demonstrates the main drawback with testing values: unless you get different outcomes (e.g., in one case, Quantity A is greater, and in another case, Quantity B is greater), you can't be certain of the correct answer.

For example, let's say we have the following question:

Given: x > 0
Quantity A: x
Quantity B: x²


case i: Since x > 0, x COULD equal 2. So, we get:
Quantity A: 2
Quantity B: 2² = 4
In this case, Quantity B is greater

case ii: Since x > 0, x COULD equal 3. So, we get:
Quantity A: 3
Quantity B: 3² = 9
In this case, Quantity B is greater

So, after testing two possible x-values, can we conclude that Quantity B is ALWAYS greater and that the correct answer is B?
No.
The problem is that we've tested only two possible x-values. So, it COULD be the case that we just haven't tested an x-value that yields a different outcome.

In fact, if x = 1, then the two quantities are equal
And, if x = 1/2, then Quantity A is greater

For more on this drawback of testing values, start watching the following video at 2:50
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Re: x<0 [#permalink]
GreenlightTestPrep wrote:
Asmakan wrote:
\(x<0\)

Quantity A
Quantity B
\((4)^x\)
\((5^{x+1})\)




Useful property #1: \(\frac{x^k}{y^k}=(x/y)^k\)

Useful property #2: \((x/y)^-k=(y/x)^-k\)

Given:
Quantity A: \((4)^x\)

Quantity B: \(5^{x+1}\)


Rewrite Quantity B as follow:
Quantity A: \(4^x\)

Quantity B: \((5^x)(5^1)\)


Divide both quantities by \(5^x\) to get:
Quantity A: \(\frac{4^x}{5^x}\)

Quantity B: \(5^1\)


Simplify both quantities to get:
Quantity A: \((\frac{4}{5})^x\)

Quantity B: \(5\)


Since we're told that x < 0, \((\frac{4}{5})^x\) can become infinitely large. Here's what I mean.

If x = -1, then \((\frac{4}{5})^x=(\frac{4}{5})^{-1}=(\frac{5}{4})^{1}=\frac{5}{4}=1.25\)

If x = -2, then \((\frac{4}{5})^x=(\frac{4}{5})^{-2}=(\frac{5}{4})^{2}=\frac{25}{16}≈1.56\)

If x = -3, then \((\frac{4}{5})^x=(\frac{4}{5})^{-3}=(\frac{5}{4})^{3}=\frac{125}{64}≈1.95\)

As we can see, \((\frac{4}{5})^x\) can become super big.
For example,
If x = -20, then \((\frac{4}{5})^x=(\frac{4}{5})^{-20}=(\frac{5}{4})^{20}≈86.7\)

Since Quantity A can range from 1.25 to infinity, the correct answer is D.

Cheers,
Brent




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thank you!
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Re: x<0 [#permalink]
thank you!
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