If \(3^k + 3^k = (3^9)^{3^9}-3^k\), then what is the value of k?

(A) 11/3
(B) 11/2
(C) 242
(D) 3^10
(E) 3^11 – 1

Kudos for a correct solution.
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Hi All,

The answer choices to this question are written in a way that helps us to avoid some of the math involved.

We're told that 3^K + 3^K = (3^9)^[3^9] - 3^K. We're asked for the value of K.

To start, we should move all like terms to one side...

3^K + 3^K + 3^K = (3^9)^[3^9]

The 'left side' can be rewritten....

3(3^K) = (3^9)^[3^9]

3^(K+1) = (3^9)^[3^9]

Both sides have the same "base 3", so since the "right side exponent" is clearly a BIG INTEGER, we know that...

K+1 = a BIG INTEGER
K = a BIG INTEGER - 1

There's only one answer that fits...

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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MANHATTAN GMAT OFFICIAL SOLUTION:

The common term in this problem is the recurring base of 3. We will group like terms (i.e. all the terms with k on the left side, all the other powers of 3 on the right side), then simplify each power of 3 using exponent rules.

\(3^k + 3^k = (3^9)^{3^9} - 3^k\)

\(3^k + 3^k + 3^k = (3^9)^{3^9}\)

\(3(3^k) = (3^9)^{3^9}\)

\(3^{(k + 1)} = 3^{(9*3^9)}\)

\(k + 1 = 9 * 3^9\)

\(k + 1 = 3^2 * 3^9\)

\(k = 3^{11} - 1\)

The correct answer is E.
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3^k + 3^k = (3^9)^(3^9) - (3^k)

3(3^k) = (3^9)^(3^9)

3^(k+1) = 3^([3^2]*[3^9])

k+1 = 3^2 * 3^9

k+1 = 3^(2+9) = 3^11

so, k = (3^11 - 1).
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(3^9)^3^9= 3^(3^2*3^9)=3^11
now taking all k terms on one side
3^k+3^k+3^k = 3^(k+1)
comparing power as bases are same
we get k+1=3^11, k=3^11-1
E answer'

Let’s simplify the given equation:

3^k + 3^k = (3^9)^(3^9) - 3^k

3^k + 3^k + 3^k = 3^(9 * 3^9)

Pull out the common factor 3^k from each term on the left side of the equation:

3^k * (1 + 1 + 1) = 3^(3^2 * 3^9)

3^k * (3) = 3^(3^11)

Note that the left side is now 3^k * 3^1, so we combine (add) the exponents:

3^(k + 1) = 3^(3^11)

k + 1 = 3^11

k = 3^11 - 1

Answer: E
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I think you mean how is \(3*3^k=3^{k+1}\)

Operations involving the same bases:
Keep the base, add or subtract the exponent (add for multiplication, subtract for division)
\(a^n*a^m=a^{n+m}\)

\(\frac{a^n}{a^m}=a^{n-m}\)


So, \(3*3^k=3^1*3^k=3^{k+1}\)

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