If 4^n 17^19, what is the smallest possible integer value of n ?

Question

If  4^n > 17^{19} , what is the smallest possible integer value of n?

(A) 31
(B) 33
(C) 35
(D) 37
(E) 39

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nick1816 · Accepted Answer

4^n > 17^{19}

(4^2)^{\frac{n}{2}} > 17^{19}

(16)^{\frac{n}{2}} > 17^{19}

As 16<17, n/2 must be greater than 19.

n/2 >19

n>38

E

fireagablast · Answer

4^n = (4^2)^n/2 = 16^n/2

16^n/2 > (17)^19
16^n/2 > (16+1)^19
n/2>19

only option greater for n/2 greater than 19 is 39. E

sachinsavailable · Answer

4^n>17^{19)

4^2^n/2> 17^{19}

16^n/2> 17^{19}

n/2> 19

n>38

hence E

GMATinsight · Answer

To compare the bases, powers must be made same or almost same OR to compare the powers, the bases must be same or almost same

4^n > 17^{19}

i.e.  (4^2)^{n/2} > 17^{19}

i.e.  (16)^{n/2} > 17^{19}

But since base to the left is smaller which the number is bigger so the Power on left must be greater than the power on the right

i.e. (n/2) > 19

i.e. n > 38

Answer: Option E

ScottTargetTestPrep · Answer

We can rewrite the inequality as:

16^(n/2) > 17^19

Since 16 < 17, the exponent of 16 must be greater than that of 17 in order for 16^(n/2) > 17^19. In other words, we need:

n/2 > 19

n > 38

Since 39 is the only value in the given choices that is greater than 38, choice E must be the correct answer.

Answer: E

CareerLauncherIE · Answer

Let's understand this better:-
Here is the solution:

We need n such that:-
4^n > 17^19

Take logs:
n.log⁡4 > 19.log⁡17
n > 19.log⁡17/log⁡4

Using powers of 2, we can see that 2^38 < 17^19 < 2^39

So the smallest integer n is 39.
Answer: 39

If 4^n > 17^19, what is the smallest possible integer value of n ?

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