population manhattan word problem

[viny]

The population of grasshoppers doubles in a particular field every year. Approximately how many years will it take the population to grow from 2,000 grasshoppers to 1,000,000 or more?

i am trying to solve it using geometric progression where b=a *(r^(n-1)) :where b is the nth term and a is the first
i am getting 1000000=2000(2^n-1)
which on solving gives me 2^(n-1)=500
i want to know how should i convert 500 to the power of 2 inorder to easily solve for n.what i did basically is i know 2^9 is 512 which is approximately equal to 500 so i considered my new eqn to be
2^n-1 = 2^9 which gives me n value to be 10.
but the real answer is 9

please help me on where i am going wrong and also a better way to convert a number to the nearest exponent like here i converted 500 in power of 2 as 2^9 ,is there any method to do the same.

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Hi viny,

There are a couple of organizational issues that I think are making this question MUCH harder than it needs to be...

1) Why go through all the trouble of setting up and solving a complex math approach when you can literally "count doubles" and have the correct answer in under 30 seconds? When in doubt, just go for whatever approach is easiest.

2) Since you're starting at 2,000 grasshoppers, your first term is 2,000 BUT that's when n = 1. N=1 is NOT the first year, it's the "zero year." So, N= 2 is the second term, but 1 year into the process. N = 3 is the third term, but 2 years into the process, etc. This is why you thought that n = 10 meant 10 years, when it actually means 9 years.

3) Maybe you used an = sign for organizational purposes, but the prompt says 1,000,000 OR MORE, so you should have an inequality, not an equation.

GMAT assassins aren't born, they're made,
Rich

[Brent@GMATPrepNow]

The population of grasshoppers doubles in a particular field every year. Approximately how many years will it take the population to grow from 2,000 grasshoppers to 1,000,000 or more?

i am trying to solve it using geometric progression where b=a *(r^(n-1)) :where b is the nth term and a is the first
i am getting 1000000=2000(2^n-1)
which on solving gives me 2^(n-1)=500
i want to know how should i convert 500 to the power of 2 in order to easily solve for n.what i did basically is i know 2^9 is 512 which is approximately equal to 500 so i considered my new eqn to be
2^n-1 = 2^9 which gives me n value to be 10.
but the real answer is 9

please help me on where i am going wrong and also a better way to convert a number to the nearest exponent like here i converted 500 in power of 2 as 2^9 ,is there any method to do the same.

It should be 2000(2^n) not 2000(2^n-1).
To see why, let's draw a table and look for a pattern.

Year 0: population = 2000
Year 1: population = 2000(2)
Year 2: population = 2000(2)(2) = 2000(2^2)
Year 3: population = 2000(2)(2)(2) = 2000(2^3)
Year 4: population = 2000(2)(2)(2)(2) = 2000(2^4)
.
.
.
Year n: population = 2000(2^n)

So, we get the following equation: 2000(2^n) = 1,000,000
Divide both sides by 2000 to get 2^n = 500
2^8 = 256 (nope)
2^9 = 512 (bingo)

n = 9

Cheers,
Brent