Dear Christophestanic,
I'm happy to respond.

I gather that you are new to GMAT Club. First, I will give you some advice.
1) This particular subforum, the "Ask GMAT Expert" forum, is for asking general study questions (study plans, book reviews, etc.). For a particular math or verbal content question, you should post in the math or verbal forums, not here.
2) VERY IMPORTANT: Never start a brand new thread for a math question before searching for it. Any GMAT practice math question from virtually any source has already been posted here and discussed in depth. It adds unnecessarily to the complexity of this site to start a brand new thread on a question that has already been discussed. More importantly, you probably can find your answer already here, in the discussions in these pre-existent threads. For example, I found this math question here:
if-x-5-25-and-x-x-5-k-what-is-k-190765.html

Please keep all that in mind when you post in the future.

Now, as to your math question: what's a little embarrassing for Veritas is that the answer they give on that screenshot is not correct. The variable k does not equal 27 ---- rather it equals k = 5^27. Actually, there are a few mistakes on that screenshot. Usually, Veritas is a little more solid than this.

Rather than explain their version, which has multiple mistakes, let's just start from scratch.

x = 5^25, a big number

x^x = (5^25)^(5^25), a super-big number

To simplify this, I am going to call the first 25 A, and the final part in parentheses B, so A = 25, and B = (5^25)

Then,
x^x = (5^A)^B

In this form, you should recognize it as one of the laws of exponents: a power to an exponent means multiply the exponents.
x^x = (5^A)^B = 5^(AB) = 5^k

Thus, k is simply the product of what I called A and B.

K = A*B = (25)*(5^25) = (5^2)*(5^25) = 5^(2 + 25) = 5^27

Notice we used the "product of the powers" law of exponents there. Once again, notice that k does NOT equal 27, as claimed on that screenshot. Instead, k = 5^27.

BTW, you don't need to know this, but some cool big number facts:
k = 5^27 = 7.45 x 10^18, more than a billion squared.

x^x =5^k would be a number with more than 5.2 x 10^18 digits. We would need a piece of paper larger than our entire galaxy simply to write this number down!

Here are some more challenging exponent problems for practice:
http://magoosh.com/gmat/2014/challengin ... and-roots/

Does all this make sense?
Mike
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x=5^25
=> x^x = (5^25)^(5^25 ) = 5^(25 * 5^25 ) [because (x^y)^z = x*yz)]
= 5^(5^2 * 5^25) = 5^(5^(2+25)) [be]cause x^a * x^b = x^(a+b)
= 5^(5^27)

Given that
x^x=5^k
=> 5^(5^27) = 5^k
Since the base is same so the power will also be same
=> k = 5^27

So, answer will be B

Hope it helps!

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F x=5^25 and x^x=5^k, what is k

a)5^26 b)5^27 c)5^50 d)5^52 e) 5^625


Since x=5^25
x^x=[5^25]^(5^25)
=5^[(5^2)*(5^25)]
=5^(5^27)

Answer B
I am not able type the power properly if someone could help with this.

Similar question to practice: new-tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029232
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See the file attached. Does anyone know how we pass from 5^25 to 25 ?

mikemcgarry, thank you for talking about the guidelines for posting.

Christophestanic, please follow posting guidelines (rules-for-posting-please-read-this-before-posting-133935.html).

Also, do not use pictures to stand for actual questions.

As per the guidelines, type in the complete question with 5 options and OA for the same. This is very important for initiating a discussion on that particular question.

Moved to PS forum.
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x = 5^25
x^x = (5^25)^(5^25)
= (5^(5^2))^(5^25)
= (5)^(5^2 * 5^25)
= 5^(5^27)

Answer B
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Can we use logarithms if we know them?

Posted from my mobile device

BodhiChatterjee

Yes you can

x = \(5^{25}\)
=> log x = 25 log 5

and \(x^x\) = \(5^k\)
=> x log x = k log 5
=> x * 25 log 5 = k log 5
=> 25x = k
=> k = 25x = 25 * \(5^{25}\)= \(5^2 * 5^{25}\) = \(5^{2+25}\) = \(5^{27}\)

So, answer will be B
Hope it helps!
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