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# If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?

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If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?  [#permalink]

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06 Mar 2017, 01:49
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If $$4^a∗2^b=16^4$$, what is the minimum possible value of (32–ab)?

A. 32
B. 16
C. 8
D. 0
E. -16

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Re: If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?  [#permalink]

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07 Mar 2017, 21:39
5
1
Bunuel wrote:
If $$4^a∗2^b=16^4$$, what is the minimum possible value of (32–ab)?

A. 32
B. 16
C. 8
D. 0
E. -16

$$4^a * 2^b = 16^4 \iff 2^{2a} * 2^b = 2 ^ 16 \iff 2a+b=16$$

If either $$a$$ or $$b$$ is negative, then $$ab$$ is negative. Hence $$32-ab > 32$$.

If either $$a$$ or $$b$$ equals to 0, then $$32-ab=32$$

If both $$a$$ and $$b$$ is positive, using AM-GM inequality, we have

$$16=2a+b \geq 2 \sqrt{2a*b} \implies 8 \geq \sqrt{2ab} \implies ab \leq 32$$.

$$ab=32 \iff 2a=b=8 \iff a=4$$ and $$b=8$$.

Hence, $$32 - ab \geq 32-32=0$$. The answer is D
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Re: If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?  [#permalink]

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06 Mar 2017, 07:51
1
Bunuel wrote:
If $$4^a∗2^b=16^4$$, what is the minimum possible value of (32–ab)?

A. 32
B. 16
C. 8
D. 0
E. -16

$$4^a∗2^b=16^4$$ -> 2a+b=16. When a=4 and b=8, ab=32 hence the lowest value of (32-ab) is 0.

Hence I'm getting D.
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Re: If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?  [#permalink]

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07 Mar 2017, 21:13
Diwakar003 wrote:
Bunuel wrote:
If $$4^a∗2^b=16^4$$, what is the minimum possible value of (32–ab)?

A. 32
B. 16
C. 8
D. 0
E. -16

$$4^a∗2^b=16^4$$ -> 2a+b=16. When a=4 and b=8, ab=32 hence the lowest value of (32-ab) is 0.

Hence I'm getting D.

Okay, but how can you be absolutely sure that that's the absolute lowest possible?
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Re: If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?  [#permalink]

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07 Mar 2017, 21:43
nguyendinhtuong wrote:
Bunuel wrote:
If $$4^a∗2^b=16^4$$, what is the minimum possible value of (32–ab)?

A. 32
B. 16
C. 8
D. 0
E. -16

$$4^a * 2^b = 16^4 \iff 2^{2a} * 2^b = 2 ^ 16 \iff 2a+b=16$$

If either $$a$$ or $$b$$ is negative, then $$ab$$ is negative. Hence $$32-ab > 32$$.

If either $$a$$ or $$b$$ equals to 0, then $$32-ab=32$$

If both $$a$$ and $$b$$ is positive, using AM-GM inequality, we have

$$16=2a+b \geq 2 \sqrt{2a*b} \implies 8 \geq \sqrt{2ab} \implies ab \leq 32$$.

$$ab=32 \iff 2a=b=8 \iff a=4$$ and $$b=8$$.

Hence, $$32 - ab \geq 32-32=0$$. The answer is D

Fancy.

Note that AM-GM is pretty firmly beyond the scope of necessary GMAT tools, but this does work.

However, also note that there is a way to prove the minimum without resorting to this tool...
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Re: If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?  [#permalink]

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07 Mar 2017, 23:32
AnthonyRitz wrote:
nguyendinhtuong wrote:
Bunuel wrote:
If $$4^a∗2^b=16^4$$, what is the minimum possible value of (32–ab)?

A. 32
B. 16
C. 8
D. 0
E. -16

$$4^a * 2^b = 16^4 \iff 2^{2a} * 2^b = 2 ^ 16 \iff 2a+b=16$$

If either $$a$$ or $$b$$ is negative, then $$ab$$ is negative. Hence $$32-ab > 32$$.

If either $$a$$ or $$b$$ equals to 0, then $$32-ab=32$$

If both $$a$$ and $$b$$ is positive, using AM-GM inequality, we have

$$16=2a+b \geq 2 \sqrt{2a*b} \implies 8 \geq \sqrt{2ab} \implies ab \leq 32$$.

$$ab=32 \iff 2a=b=8 \iff a=4$$ and $$b=8$$.

Hence, $$32 - ab \geq 32-32=0$$. The answer is D

Fancy.

Note that AM-GM is pretty firmly beyond the scope of necessary GMAT tools, but this does work.

However, also note that there is a way to prove the minimum without resorting to this tool...

AnthonyRitz please explain another way
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If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?  [#permalink]

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Updated on: 07 Mar 2017, 23:54
4
mohshu wrote:
AnthonyRitz please explain another way

Fair enough. Try this:

$$2a+b=16$$

$$32-ab = 32-a(16-2a) = 32-16a+2a^2 = 2(a^2-8a+16) = 2(a-4)^2$$

Since $$(a-4)^2 \geq 0$$ with equality when $$a = 4$$, the minimum is $$0$$ when $$a = 4$$ and $$b = 8$$.

Originally posted by AnthonyRitz on 07 Mar 2017, 23:39.
Last edited by AnthonyRitz on 07 Mar 2017, 23:54, edited 2 times in total.
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If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?  [#permalink]

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07 Mar 2017, 23:46
AnthonyRitz wrote:
nguyendinhtuong wrote:
Bunuel wrote:
If $$4^a∗2^b=16^4$$, what is the minimum possible value of (32–ab)?

A. 32
B. 16
C. 8
D. 0
E. -16

$$4^a * 2^b = 16^4 \iff 2^{2a} * 2^b = 2 ^ 16 \iff 2a+b=16$$

If either $$a$$ or $$b$$ is negative, then $$ab$$ is negative. Hence $$32-ab > 32$$.

If either $$a$$ or $$b$$ equals to 0, then $$32-ab=32$$

If both $$a$$ and $$b$$ is positive, using AM-GM inequality, we have

$$16=2a+b \geq 2 \sqrt{2a*b} \implies 8 \geq \sqrt{2ab} \implies ab \leq 32$$.

$$ab=32 \iff 2a=b=8 \iff a=4$$ and $$b=8$$.

Hence, $$32 - ab \geq 32-32=0$$. The answer is D

Fancy.

Note that AM-GM is pretty firmly beyond the scope of necessary GMAT tools, but this does work.

However, also note that there is a way to prove the minimum without resorting to this tool...

There is another way.

$$(2a)+b=16 \implies ((2a)+b)^2=16^2 \implies (2a)^2 + b^2 + 2*(2a)*b=16^2$$
$$(2a)^2 + b^2 - 2*(2a)*b = 16^2 - 4*(2a)*b \implies (2a-b)^2 = 16^2 - 4*(2a)*b$$

We have $$(2a-b)^2 \geq 0 \;\; \forall a,b \in R$$.
Hence $$16^2 -4 *(2a)*b \geq 0 \implies 4*(2a)*b \leq 16^2 \implies 8ab \leq 16^2 \implies ab\leq 32$$
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Re: If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?  [#permalink]

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08 Mar 2017, 01:58
LHS: 4^a x 2^b = 2^(2a+b)
RHS: 16^4 = 2^16

hence 2a+b =16 or b = 16-2a
putting in the expression 32-ab we get 32-a(16-2a) = 32-16a +2a^2

minima of quadratic equation comes at 16/4 = 4.

for a = 4 b = 8. so ab = 32.hence 32-ab = 32-32 = 0.

Option D
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Re: If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?  [#permalink]

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09 Mar 2017, 17:15
1
Bunuel wrote:
If $$4^a∗2^b=16^4$$, what is the minimum possible value of (32–ab)?

A. 32
B. 16
C. 8
D. 0
E. -16

We can simplify the given equation by re-expressing 4 as 2^2 and 16 as 2^4:

(2^2)^a * 2^b = (2^4)^4

2^(2a) * 2^b = 2^16

2^(2a + b) = 2^16

In an equation, when the bases are the same, the exponents must be equal:

2a + b = 16

b = 16 - 2a

We need to find the minimum value of (32 - ab). We have determined that b = 16 - 2a, so we need to find the minimum value of 32 - a(16 - 2a).

Let’s simplify this expression:

32 - 16a + 2a^2

2a^2 - 16a + 32

The above is a quadratic expression. Recall that the graph of y = ax^2 + bx + c is a parabola. It opens up when a > 0, and its vertex will be the minimum point. To find the x-value of the vertex, we can use the formula x = -b/(2a). As for the minimum value of the quadratic expression (i.e., the y value), we can plug the x-value of the vertex back into the expression.

Thus, the minimum value of the expression occurs when a = -(-16)/[2(2)] = 16/4 = 4, and the minimum value is:

2(4)^2 - 16(4) + 32

32 - 64 + 32 = 0

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Re: If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?  [#permalink]

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20 Apr 2017, 13:39
Took a little more time but solved it
Choice is: D
here you go
4^a*2^b=16^4
2^2a*2^b=2^16
so
2a+b=16
option E ruled out min we get is Zero
So check
2(4)+8=16
LHS=RHS
put in 32-ab
32-(4)(8)
0
Here is our answer
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Re: If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?  [#permalink]

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11 Sep 2018, 09:33
Well, once you have the equation 2a+b=16, equate answer choices to "32-ab".
I tried with choice D - and it satisfied both eqns.
The only other answer choice lower than 0 was E, which does not satisfy 2a+b=16.
So we're left with D.
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Re: If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?  [#permalink]

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17 Sep 2018, 07:58
I think I went about this the wrong way but I ended up with the right answer:

$$4^a * 2^b = 16^4$$
$$4^a * 2^b = (2 * 2 * 4)^4$$
$$4^a * 2^b = (2^2 * 4)^4$$
$$4^a * 2^b = 2^8 * 4^4$$

So I took $$a = 4$$ and $$b = 8$$ and plugged it into $$(32 - ab)$$ and got 0. D.
I didn't feel totally confident about it though.
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Re: If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)?  [#permalink]

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23 Sep 2018, 03:32
broall wrote:
Bunuel wrote:
If $$4^a∗2^b=16^4$$, what is the minimum possible value of (32–ab)?

A. 32
B. 16
C. 8
D. 0
E. -16

$$4^a * 2^b = 16^4 \iff 2^{2a} * 2^b = 2 ^ 16 \iff 2a+b=16$$

If either $$a$$ or $$b$$ is negative, then $$ab$$ is negative. Hence $$32-ab > 32$$.

If either $$a$$ or $$b$$ equals to 0, then $$32-ab=32$$

If both $$a$$ and $$b$$ is positive, using AM-GM inequality, we have

$$16=2a+b \geq 2 \sqrt{2a*b} \implies 8 \geq \sqrt{2ab} \implies ab \leq 32$$.

$$ab=32 \iff 2a=b=8 \iff a=4$$ and $$b=8$$.

Hence, $$32 - ab \geq 32-32=0$$. The answer is D

Can someone pls explain the AM-GM theory here. Thanks a ton in advance
Re: If 4^a∗2^b=16^4, what is the minimum possible value of (32–ab)? &nbs [#permalink] 23 Sep 2018, 03:32
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