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Re: If 4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^24, what is the value of x? [#permalink]
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Bunuel wrote:
If \(4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^{24}\), what is the value of x?

(A) 3
(B) 5
(C) 6
(D) 8.5
(E) 11.5


Kudos for a correct solution.


Now, 4^2x can be written as 2^2(2x)=2^4x
Therefore, (4)2^4x=2^48
2^4x+2=2^48
4x+2=48
4x=46
x=11.5
Answer E
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Re: If 4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^24, what is the value of x? [#permalink]
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The correct answer is E.

Converting to common base of 2, we get 2^4x + 2^4x + 2^4x + 2^4x = 2^48

> Take 2^4x common: 2^4x (1+1+1+1) = 2^48

>2^4x (4) = 2^48

> 2^4x +2 = 2^48

> 4x + 2 = 48

> 4x = 46

> x = 11.5
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Re: If 4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^24, what is the value of x? [#permalink]
Expert Reply
Bunuel wrote:
If \(4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^{24}\), what is the value of x?

(A) 3
(B) 5
(C) 6
(D) 8.5
(E) 11.5


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Here, you can note that you do a few things quite well with exponential terms:

Break their bases down to primes to get common bases. Multiply them.

So when you see a problem like this, you should recognize your strengths with exponents and look to rearrange the algebra to take advantage of them. Breaking the 4 terms down to prime factors (2), you get:

\((2^2)^{2x} + 2^{4x} + (2^2)^{2x} + 2^{4x} = (2^2)^{24}\)

Then you can get back to multiplication to eliminate the parentheses:

\(2^{4x} + 2^{4x} + 2^{4x} + 2^{4x} = 2^{48}\)

Again, look for chances to do what you do well – and you know that if you can multiply the terms on the left instead of adding them, you’re then multiplying exponential terms with a common base…that’s your strength. In this problem, you may recognize quickly that you have four of the same term, and can express it as:

\(4(2^{4x}) = 2^{48}\)

Were the problem slightly more difficult, or you didn’t make that recognition, you might need to factor out the common exponential term so that you can multiply it that way:

\(2^{4x}(1+1+1+1)=2^{48}\)

\(2^{4x}(4) = 2^{48}\)

Either way, you end up with the same multiplication, which is what’s most important – now you’re doing what you do well.

\(4(2^{4x}) = 2^{48}\)

One more step is to, again, break down different bases into primes so that you can again multiply exponents. 4 = 2^2, so you have:

\(2^2(2^{4x}) = 2^{48}\)

And because you’re pretty quick when multiplying exponents of the same base, you should recognize that that can be expressed as:

\(2^{4x+2} = 2^{48}\)

Now that the bases are the same and the terms are set equal, you can note that:

\(4x+2 = 48\)

\(4x = 46\)

\(x = 11.5\), and the answer is E.
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Re: If 4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^24, what is the value of x? [#permalink]
Bunuel wrote:
If \(4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^{24}\), what is the value of x?

(A) 3
(B) 5
(C) 6
(D) 8.5
(E) 11.5


Kudos for a correct solution.


\(4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^{24}\)

Or, \(2^{4x} + 2^{4x} + 2^{4x} + 2^{4x} = 2^{48}\)

Or, \(2^{4x} ( 1 + 1 + 1 + 1 ) = 2^{48}\)

Or, \(2^{4x} *2^2 = 2^{48}\)

Or, \(2^{4x} = 2^{46}\)

Or, \(4x = 48\)

Or, \(x = 11.5\)

Thus, answer must be (E) 11.50
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Re: If 4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^24, what is the value of x? [#permalink]
Expert Reply
Bunuel wrote:
If \(4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^{24}\), what is the value of x?

(A) 3
(B) 5
(C) 6
(D) 8.5
(E) 11.5


Let’s simplify the given equation:

4^2x + 2^4x + 4^2x + 2^4x = 4^24

2^4x + 2^4x + 2^4x + 2^4x = 2^48

2^4x(1 + 1 + 1 + 1) = 2^48

2^4x(4) = 2^48

2^4x(2^2) = 2^48

2^(4x + 2) = 2^48

4x + 2 = 48

4x = 46

x = 46/4 = 23/2 = 11.5

Answer: E
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Re: If 4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^24, what is the value of x? [#permalink]
First Solution post on GMAT Club
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Simplify and get
2^4x(2^2) = 2^48
Simplify this and get
4x+2 = 48
Simplify this and get 11.5


Bunuel wrote:
If \(4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^{24}\), what is the value of x?

(A) 3
(B) 5
(C) 6
(D) 8.5
(E) 11.5


Kudos for a correct solution.
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Re: If 4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^24, what is the value of x? [#permalink]
Expert Reply
\(4^{2x}\) + \(2^{4x}\) + \(4^{2x}\) + \(2^{4x}\) = \(4^{24}\)

=> \(4^{2x}\) + \(4^{2x}\) + \(2^{4x}\) + \(2^{4x}\) = \(4^{24}\)

=> \(((2)^2)^ {2x}\) + \(((2)^2)^ {2x}\) + \(2^{4x}\) + \(2^{4x}\) = \(((2)^2)^{24}\)

=> \(2^{4x}\) + \(2^{4x}\) + \(2^{4x}\) + \(2^{4x}\) = \(2^{48}\)

=> \(2^{4x}\) * 4 = \(2^{48}\)

=> \(2^{4x}\) = \(\frac{2^{48} }{ 2^2}\)

=> \(2^{4x}\) = \(2^{46}\)

Comparing bases,

=> 4x = 46

=> x = 11.5

Answer E
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Re: If 4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^24, what is the value of x? [#permalink]
Asked: If \(4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^{24}\), what is the value of x?

\(2^{4x} + 2^{4x} + 2^{4x} + 2^{4x} = 2^{48}\)
\(2^{4x+2} = 2^{48}\)
4x + 2 = 48
4x = 46
x = 11.5

IMO E
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Re: If 4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^24, what is the value of x? [#permalink]
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Re: If 4^{2x} + 2^{4x} + 4^{2x} + 2^{4x} = 4^24, what is the value of x? [#permalink]
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