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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, 00:49
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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, 06:51
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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 (32ab) 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, 20:13
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 (32ab) 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, 20:39
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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 \(32ab > 32\). If either \(a\) or \(b\) equals to 0, then \(32ab=32\) If both \(a\) and \(b\) is positive, using AMGM 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, \(32  ab \geq 3232=0\). 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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07 Mar 2017, 20:43
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 \(32ab > 32\). If either \(a\) or \(b\) equals to 0, then \(32ab=32\) If both \(a\) and \(b\) is positive, using AMGM 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, \(32  ab \geq 3232=0\). The answer is D Fancy. Note that AMGM 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, 22:32
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 \(32ab > 32\). If either \(a\) or \(b\) equals to 0, then \(32ab=32\) If both \(a\) and \(b\) is positive, using AMGM 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, \(32  ab \geq 3232=0\). The answer is D Fancy. Note that AMGM 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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07 Mar 2017, 22:39
mohshu wrote: AnthonyRitz please explain another way Fair enough. Try this: \(2a+b=16\) \(32ab = 32a(162a) = 3216a+2a^2 = 2(a^28a+16) = 2(a4)^2\) Since \((a4)^2 \geq 0\) with equality when \(a = 4\), the minimum is \(0\) when \(a = 4\) and \(b = 8\).
Last edited by AnthonyRitz on 07 Mar 2017, 22:54, 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, 22:46
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 \(32ab > 32\). If either \(a\) or \(b\) equals to 0, then \(32ab=32\) If both \(a\) and \(b\) is positive, using AMGM 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, \(32  ab \geq 3232=0\). The answer is D Fancy. Note that AMGM 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. \((2a)+b=16 \implies ((2a)+b)^2=16^2 \implies (2a)^2 + b^2 + 2*(2a)*b=16^2\) \((2a)^2 + b^2  2*(2a)*b = 16^2  4*(2a)*b \implies (2ab)^2 = 16^2  4*(2a)*b\) We have \((2ab)^2 \geq 0 \;\; \forall a,b \in R\). 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, 00:58
LHS: 4^a x 2^b = 2^(2a+b) RHS: 16^4 = 2^16
hence 2a+b =16 or b = 162a putting in the expression 32ab we get 32a(162a) = 3216a +2a^2
minima of quadratic equation comes at 16/4 = 4.
for a = 4 b = 8. so ab = 32.hence 32ab = 3232 = 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, 16:15
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 reexpressing 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 xvalue 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 xvalue 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 Answer: 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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20 Apr 2017, 12:39
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 32ab 32(4)(8) 0 Here is our answer
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