If (5/4)^(-n)

Question

If  (\frac{5}{4})^{-n} < 16^{-1} . What is the least integer value of n?

A. 12
B. 13
C. 14
D. 15
E. 16

chetan2u · Accepted Answer

if you counter such Q, an approximation way would be..

(\frac{5}{4})^{-n} < 16^{-1}..........(\frac{4*2}{5*2})^n<\frac{1}{16}......2^{3n}*16<10^n....2^{3n+4}<10^n

now 2^{10}=1024  and  10^3 = 1000  ..nearly equal so lets convert ..

If you could get some even higher close values, the answer will be even closer   
 1024^{\frac{3n+4}{10}}<1000^{\frac{n}{3}} ...

so  \frac{3n+4}{10}<\frac{n}{3}........9n+12<10n....n>12 
 next integer value after 12 is 13.... 
so ans 13
B

hellosanthosh2k2 · Answer

(4/5)^n < 1/16

as both 4/5 and 1/16 are positive

we take 4th root on both sides

(4/5)^{n/4} < 1/2

let  k = n/4

when k = 3,  (4/5)^3  =  64/125  is slightly greater than 1/2,

so least value of n for which  (4/5)^n < 1/16  is n = 3(4) + 1 = 13

shashankism · Answer

The options are so close, How can one say that on taking 1024 ~ 1000 , one will get correct answer.. 
one might get n>13 or n>11 instead of n>12 in some other question of this type....

We need to find some other approach..

newfolder · Answer

I would say taking logs on both sides would be better.

On taking logs, we get

(n+2)*log4 < n*log 5

=> (n+2)/n < log5/log4

Now log5 = 1-log2 = 1-.301 =.699
log4 = 2*log2 =0.602

=> (n+2)/n = 0.699/0.602

Putting n=12, we know that the inequality is not satisfied. Because LHS = 7/6 whereas RHS is less than 7/6 (0.699/0.602 is less than 0.7/0.6). Hence next integral value of n is correct

niks18 · Answer

The question stem can be written as  (\frac{4}{5})^n < \frac{1}{16}

Now we know that  16 = 2^4  so if we can make  (\frac{4}{5})  some approximate form of  2 , then our job might become easier.

(\frac{4}{5})^3 = \frac{64}{125}  this is slightly than greater  \frac{1}{2}

Hence  (\frac{4}{5})^3 = \frac{1}{2} ,

and  (\frac{1}{2})^4 = \frac{1}{16}

Substitute the value of  \frac{1}{2}  to get  (\frac{4}{5})^{12}  but this will be greater than  \frac{1}{16} .

Hence to make it lower we will have to multiply  (\frac{4}{5})^{12}  by a lower value  \frac{4}{5}

Therefore  (\frac{4}{5})^{13}<\frac{1}{16}

So  n=13

Option  B

P.S: This is a very tough question and probably beyond the scope of GMAT. Unless you have solved similar questions, it will be very difficult to solve it in real GMAT within 2 minutes.

Bunuel  Do you have repository of similar kind of questions?

cbh · Answer

One method (gmat style

) (5/4)^(−n)<16^(−1) (4/5)^n<1/2^4 so left side should be less than 1/2 we raise to cube (4/5)^3=64/125 and multiply right side by 64 just to compare, makes 64/128, so the inequality is still not satisfied but just a bit, it's clear that if multiply the left side by 4/5 one more time the inequality will be respected But as we have 4 power on the right 1/2^4 then for the left (4/5)^12 *4/5 which is 13

cbh · Answer

And another approach for math geeks, solving through logs, I always recommend to use ln (natural log) just memorize 6 numbers by heart (ln for 2 3 5 6 7 and 10) you will cove most of that kind of questions https://gmatclub.com/forum/which-of-the-following-is-the-largest-163644.html#p1930561
(5/4)^(−n)<16^(−1)
ln (4/5)^(n)< ln 2^(−4)
as per log property
n *ln (4/5)< -4 * ln 2
another property
n *(ln 4 - ln5) < -4 * ln 2
n* (1.4 - 1.6) < - 4 * 0.7
n * (-0.2) < - 2.8
then
n < - 2.8/ -0.2 =>> n < 14 so 13

OC2910 · Answer

mikemcgarry  can you please explain this Magoosh question?

mikemcgarry · Answer

Dear StrugglingGmat2910 , I'm happy to respond.

My friend, this was posted as a Magoosh question, but I don't believe it's one that we still have in our Product. This is a very hard question, one that is probably best solved with math that is completely beyond the GMAT. I am going to say that you don't need to worry about this question at all for the GMAT. Does this make sense? Mike

OC2910 · Answer

thank you sir for your reply

abhishekrukhaiyar · Answer

(\frac{5}{4})^{-n} < 16^{-1} => A = (\frac{4}{5})^n*2^4 < 1 For n = 0, A =16; n = 1, A = 16*0.8. So, as the value of n increases, A decreases and as n is increased by 1, A decreases by 80 %. We need to look for the value of n for which A >1 and A*80% less than 1. As the power of 2 is 4, I was looking to test multiple of 4 for n while solving. So, I decided to first test with n = 12; from the list of option. As it will bit simplify A. For n = 12, A = (\frac{4}{5})^{12}*2^4 = (\frac{4^3}{5^3})^4*2^4 = (\frac{64}{125})^4*2^4 = (\frac{128}{125})^4 A will be slightly greater than 1. But, definitely 80% of A will be just greater than 0.8. So, if n=13, value of A will be around 0.8 and it is the least integer value for which the inequality satisfies. Hence, B. I might be a bit lucky.

as the question was definitely not an easy one to crack.

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If (5/4)^(-n) < 16^(-1). What is the least integer value of n?

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