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If for some value of x, 5^x + 5^(–x) = B, then which of the following

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lalania1
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If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] New post   Updated on: 24 Nov 2016, 08:03
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If for some value of x, 5^x + 5^(–x) = B, then which of the following is equal to 25^x + 25^(–x)?

A) 5B
B) 5B - 10
C) B^2
D) B^2 - 2
E) B^2 - 10
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D

Originally posted by lalania1 on 24 Nov 2016, 07:57.
Last edited by Bunuel on 24 Nov 2016, 08:03, edited 1 time in total.
Edited the question.
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Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] New post   27 Jun 2018, 13:13
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lalania1 wrote:
If for some value of x, 5^x + 5^(–x) = B, then which of the following is equal to 25^x + 25^(–x)?

A) 5B
B) 5B - 10
C) B^2
D) B^2 - 2
E) B^2 - 10


Let x=0.

\(B = 5^x + 5¯^x = 5^0 + 5^0 = 1 + 1 = 2.\)
\(25^x + 25¯^x = 25^0 + 25^0 = 1 + 1 =\) \(2\).

When B=2 is plugged into the answer choices, the result must be the value in blue.
Only D works:
\(B^2 - 2 = 2^2 - 2 = 4 - 2 = 2\).

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D
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lalania1
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Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] New post   24 Nov 2016, 07:59
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Hi all,

5^–x = 5 to the -x exponent. I was not able to find the right way to express it. Feel free to edit with a fix if you have it.

Looking forward to your answers on this question
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Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] New post   24 Nov 2016, 08:07
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lalania1 wrote:
If for some value of x, 5^x + 5^(–x) = B, then which of the following is equal to 25^x + 25^(–x)?

A) 5B
B) 5B - 10
C) B^2
D) B^2 - 2
E) B^2 - 10


Square \(5^x + 5^{–x} = B\):

\(25^{x} +2*5^x* 5^{–x}+25^{-x} = B^2\);

\(25^{x} +2+25^{-x} = B^2\);

\(25^{x} + 25^{-x} = B^2-2\).

Answer: D.
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Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] New post   24 Nov 2016, 09:21
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lalania1 wrote:
If for some value of x, 5^x + 5^(–x) = B, then which of the following is equal to 25^x + 25^(–x)?

A) 5B
B) 5B - 10
C) B^2
D) B^2 - 2
E) B^2 - 10


Let x = 1

\(5^x + 5^{–x} = \ B \ = 5 + \frac{1}{5}\) = \(\frac{26}{5}\)

Let x = 2

\(5^x + 5^{–x} = 5^2 + \frac{1}{5^2}\) = 25 + \(\frac{1}{25}\) = \(\frac{626}{25}\)

From the above options (A) and (B) can straightaway be rejected...

(C) \(B^2 = \frac{676}{25}\) Rejected...

(D) \(B^2 - 2 = \frac{( 676 - 50 )}{25}\) = \(\frac{626}{25}\)

Hence, answer must be (D) \(\frac{626}{25}\)
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Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] New post   27 Jun 2018, 04:48
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Bunuel wrote:
lalania1 wrote:
If for some value of x, 5^x + 5^(–x) = B, then which of the following is equal to 25^x + 25^(–x)?

A) 5B
B) 5B - 10
C) B^2
D) B^2 - 2
E) B^2 - 10


Square \(5^x + 5^{–x} = B\):

\(25^{x} +2*5^x* 5^{–x}+25^{-x} = B^2\);

\(25^{x} +2+25^{-x} = B^2\);

\(25^{x} + 25^{-x} = B^2-2\).

Answer: D.


Okay I had a question, but as I typed this out I got the answer... Here is a detailed step by step of Bunuel's solution. Thanks Bunuel.

I was confused how he got the 2 \(* 5^x * 5^{-x}\), but now I see that's just \(5^0 + 5^0\)

Square \(5^x + 5^{–x} = B\)

\(B^2\)= (\(5^x + 5^{-x}\))(\(5^x + 5^{-x}\)). Foil method...

\(B^2\) = (\(5^{2x} +5^{x-x} +5^{x-x} +5^{-2x}\)) Which then would lead to

\(B^2\) = (\(5^{2x} + 5^0 + 5^0 + 5^{-2x}\)) We know \(5^0\) = 1

\(B^2\) = (\(5^{2x} + 2 + 5^{-2x}\)) Subtract the 2 to move it over to LHS. \(5^{2x} +5^{-2x} = 25^x + 25^{-x}\)

\(B^2\) -2 = \(25^x + 25^{-x}\)
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Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] New post   25 Feb 2017, 13:16
Can someone please explain why the answer is not 5B?
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Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] New post   26 Feb 2017, 03:37
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Tylcleaner wrote:
Can someone please explain why the answer is not 5B?


There are TWO solutions above:
https://gmatclub.com/forum/if-for-some- ... l#p1767286
https://gmatclub.com/forum/if-for-some- ... l#p1767321
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If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] New post   26 Feb 2017, 07:42
Answer: D

Given, 5^x + 1/ (5^x) = B
Substitute 5^x = a
a + 1/a = B

Needed = 25^x + 25^(–x) = (5^x)^2 + 1 / (5^x)2 = a^2 +1/ a^2 = (a^4 + 1) / a^2

a + 1/a = B
a^2+ 1 = a * B
a^4 + 1 + 2*(a^2) = (a^2) * (B^2) [squaring both sides]
a^4 + 1 = (a^2) * (B^2) - 2*(a^2)
(a^4 + 1) / a^2 = (B^2) - 2
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Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] New post   01 Dec 2017, 19:11
Can someone please advise where I went wrong?

I approached the question in the following manner:

Given: \(5^x+5^{-x} = B\)

Simplify: \(25^x+25^{-x} = 5^{2x}+5^{-2x} = 5(5^x+5^{-x}) = 5(B)\)

Why can't I approach like this?
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Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] New post   01 Dec 2017, 23:58
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nicoeddy wrote:
Can someone please advise where I went wrong?

I approached the question in the following manner:

Given: \(5^x+5^{-x} = B\)

Simplify: \(25^x+25^{-x} = 5^{2x}+5^{-2x} = 5(5^x+5^{-x}) = 5(B)\)

Why can't I approach like this?


If you factor 5 from \(5^{2x}+5^{-2x}\) you get \(5(5^{2x-1}+5^{-2x-1})\), not \(5(5^x+5^{-x})\).

Check below:

8. Exponents and Roots of Numbers



Check below for more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
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Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] New post   23 Apr 2018, 07:32
+1 for option D.

Use (a+b)^2=a^2+b^2+2ab.

(5^x+5^-x)^2=25^x+25^-x+2=b^2 ; The required qty is b^2-2.

Hence option D
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Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] New post   24 Apr 2018, 11:31
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lalania1 wrote:
If for some value of x, 5^x + 5^(–x) = B, then which of the following is equal to 25^x + 25^(–x)?

A) 5B
B) 5B - 10
C) B^2
D) B^2 - 2
E) B^2 - 10


Squaring both sides of the equation, we have:

[5^x + 5^(–x)]^2 = B^2

(5^2)^x + (5^2)^-x + 2(5^x)(5^-x) = B^2

25^x + 25^-x + 2(5^0) = B^2

25^x + 25^-x + 2 = B^2

25^x + 25^-x = B^2 - 2

Answer: D
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Re: If for some value of x, 5^x + 5^(x) = B, then which of the following [#permalink] New post   24 Apr 2023, 03:54
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