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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:
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3^k+3^k=(3^9)^3^9 - 3^k --> moving the - 3^k to the other side:
3(3^k) = (3^9)^3^9 --> using the properties of powers, we are now adding the exponents:
3^k+1 = 3^ (3^2)^(3^9) (here we broke 9 into 3^2)
k+1 = 3^2*3^9
k+1 = 3^11
k = 3^11 - 1.

ANS E
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Can anyone tell me what Im doing mathematically illegal.

I start by seeing all the bases are the same so I go straight to the exponents.

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

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

3k=\((3^11)\)

k=\((3^11)/3\)

k=\(3^10\)

I believe I'm doing something wrong in step 2 but not sure if it is or not can anyone confirm?
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Hi xLUCAJx,

The first error is right at the beginning - you cannot combine exponents in the way that you did:

3^K + 3^K is NOT 3^2K

3^K + 3^K = 2(3^K)

When you add 3^K to both sides, the 'left side' becomes...

3^K + 3^K + 3^K

This can be rewritten as...

3(3^K) =
(3^1)(3^K) =
3^(K+1)

Using these steps as your 'starting point', what would you do next?

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giddi77
If 3^k + 3^k = (3^9)^(3^9) - 3^k, then k = ?

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

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

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

Note that on the left side, when we add 3^k to itself three times, we have 3(3^k).

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

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

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

Since the bases are equal, we can equate the exponents:

k + 1 = 3^11

k = 3^11 - 1

Answer: E
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Bunuel
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

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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Hi ,
I was able to solve it but took about 7 mints :(.. my initial approach was :
3^k+3^k+3^k = (3^9)^(3^9)
3(3)^k =3(3^8)^(3^9)
3^k=(3^8)^(3^9)
k=8*(3^9)
I wasnt able to proceed beyond this and hence went back and re worked on it using the approach given by Brunnel. But out of curiousity, can anyone help me with where I went wrong in my approach?
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Can anyone please explain how 3^3k is transformed to 3^k+1
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Can anyone please explain how 3^3k is transformed to 3^k+1

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


8. Exponents and Roots of Numbers



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Bunuel
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

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

Is the reason why \(3^k+3^k+3^k\) doesn't equal \(3^2k\) is because exponential terms that are added or subtracted cannot be combined? I am reviewing the MGMT algebra book and didn't understand that concept at first.
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Bunuel
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

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

Is the reason why \(3^k+3^k+3^k\) doesn't equal \(3^2k\) is because exponential terms that are added or subtracted cannot be combined? I am reviewing the MGMT algebra book and didn't understand that concept at first.

Hi whollymoses,

The GMAT will sometimes test you on concepts that you know, but in ways that you are not used to thinking about. For example, I bet that you know how to do basic Arithmetic.

For example:
1 + 1 + 1 = 3(1)

In that same way, I bet that you can do basic Algebra too:
X + X + X = 3(X)

At this point, you probably recognize the pattern. Now, apply it to the math you asked about...
\(3^k+3^k+3^k = 3(3^k\))

GMAT assassins aren't born, they're made,
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whollymoses


Is the reason why \(3^k+3^k+3^k\) doesn't equal \(3^2k\) is because exponential terms that are added or subtracted cannot be combined? I am reviewing the MGMT algebra book and didn't understand that concept at first.

Yes you are correct. The only thing you can do with 3^k + 3^k + 3^k is factor out 3^k, giving you:

3^k(1 + 1 + 1) = 3^k * 3
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Bunuel, VeritasKarishma why cant we use the exponents formula of (a^x)^c = c.a^x in this question. For eg : Can this (3^9)^3^9 be written as 3^9.3^9?
Any help on this would be appreciated. Thanks
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Bunuel
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.

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

Asked: What is the value of k?

\(3.3^k = 3^{9.3^9}\)
\(3^{k+1} = 3^{3^11}\)
\(k+1 = 3^11\)
\(k = 3^11 -1\)

IMO E
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porwal1
Bunuel, VeritasKarishma why cant we use the exponents formula of (a^x)^c = c.a^x in this question. For eg : Can this (3^9)^3^9 be written as 3^9.3^9?
Any help on this would be appreciated. Thanks

The exponent rule is:

\((a^b)^c = a^{bc}\)
The exponents get multiplied.

It is not the same as \(a^b * c\)

We are given
\((3^9)^{3^9} = 3^{9*3^9}\)
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Can someone explain what I am doing wrong over here?
3^k+3^k+3^k=(3^9)^(3^9)
3^k(1+1+1)=(3^27)^9
3^k+1=3^243
k+1=242
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Varsh003
Can someone explain what I am doing wrong over here?
3^k+3^k+3^k=(3^9)^(3^9)
3^k(1+1+1)=(3^27)^9
3^k+1=3^243
k+1=242

I believe you meant to write:

\(3^k(1 + 1 + 1) = 3^k * 3 = 3^{(k + 1)}\) — you were missing brackets in your original expression.

Next, \((3^9)^{(3^9)}\) is not equal to (3^27)^9 .

Using the rule \((a^m)^n=a^{mn}\), we get:

\((3^9)^{(3^9)} = 3^{(9*3^9)}= 3^{(3^2*3^9)}= 3^{(3^{11})} \).

So we now have:

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

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

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

Hope it's clear.
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