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

Question

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

AnthonyRitz · Accepted Answer

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 $b = 8$.

broall · Answer

$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

Diwakar003 · Answer

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.

AnthonyRitz · Answer

Okay, but how can you be absolutely sure that that's the absolute lowest possible?

AnthonyRitz · Answer

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...

mohshu · Answer

AnthonyRitz  please explain another way

broall · Answer

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 $16^2 -4 *(2a)*b \geq 0 \implies 4*(2a)*b \leq 16^2 \implies 8ab \leq 16^2 \implies ab\leq 32$

KrishnakumarKA1 · Answer

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

ScottTargetTestPrep · Answer

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)/  = 16/4 = 4, and the minimum value is:

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

32 - 64 + 32 = 0

Answer: D

ehsan090 · Answer

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

anvikshashetty · Answer

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.

jameslewis · Answer

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.

Adarsh_24 · Answer

Maybe below method works too.

Since, 2a+b=16
And we need minimum of 32-ab.

32-ab is minimum when ab is maximum. Or 2ab is maximum.

If we need 2a * b maximum such that 2a + b =16
So 2a should be equal to b

4a=16 -> a=4
2a+b=16 -> b=16-2a=8

So 32-ab = 32-4*8=0

BinodBhai · Answer

New solution:  - No quadratic eqn

Given: 4^a * 2^b = 16^4

We can write 16^4 as (4*2*2)^4 = 4^4 * 2^8

So  4^a * 2^b = 4^4 * 2^8 
 Hence a=4, b=8, ab= 32

Now min value of 32-ab = 32-32=0  (D Ans)

luisdicampo · Answer

Deconstructing the Question  
We are given  4^a\cdot 2^b=16^4 .
We want the minimum possible value of  32-ab .

Key idea: rewrite everything as powers of  2  to get a linear constraint.
To minimize  32-ab , we must maximize  ab .

Step-by-step  
Rewrite in base 2:
 4^a=(2^2)^a=2^{2a} 
 16^4=(2^4)^4=2^{16}

So:
 2^{2a}\cdot 2^b=2^{16} 
which implies:
 2a+b=16 
thus:
 b=16-2a

Compute the product:
 ab=a(16-2a)=16a-2a^2

This is a concave parabola, so it is maximized at its vertex:
 a=4

Then:
 b=16-2(4)=8 
 ab=4\cdot 8=32

Therefore:
 32-ab=32-32=0

Answer: D

pappal · Answer

in order to minimize 32-ab we need to maximize ab.
given 4^a*2^b=16^4
if we take a=2 and b=12 so ab=24 
if we take a=4 and b=8 ab=32 will give the minimum value of 32-ab=0

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)?

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