mbahopeful111 wrote:This question concerns an exponential growth problem. MGMAT presents a simple chart based solution, but I wonder if the the problem can be answered through algebraic manipulation.

"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?"

Can we take the square root of both sides of the equation 1,000,000 = 2,000^(2t)? If so, why isn't the resulting expression 1,000 = 2,000^t? The correct answer is 9 years.

Thanks!

Hi mbahopeful111,

A GMAT PS question happens to be with five answer choices. Many GMAT PS questions can be answered by the help of those answer choices only, without doing any real math, and that again in a fraction of minute. Please show us the answer choices to see how. Algebra is a bad idea to crack this question. There are far faster ways already in the market. 9 is the right answer, by the way.

I agree that counting doubles would be quick.

Anyway, here's a simple way of solving it algebraically too:

2000 * 2^y = 1000000 simplifies to 2^y = 500

Therefore y = Log (500) using Log base 2

So y = 8.966


For a more detailed method using any single Log base you like:

Take logs of both sides

Log (2^y) = Log (500)

y Log (2) = Log (500) (using basic Log laws, an index inside a Log = multiplying it by the Log)

y = Log (500)/Log (2) = 8.966

Eitherway, it's straightforward high school maths.

(Perhaps I'm showing my age when I say that in the days before calculators, we all had to use logarithms to solve problems. It's a handy way of changing a curved graph into a straight line when solving more difficult real-life or science problems too.)

Mathsbuddy wrote:I agree that counting doubles would be quick.

Anyway, here's a simple way of solving it algebraically too:

2000 * 2^y = 1000000 simplifies to 2^y = 500

Therefore y = Log (500) using Log base 2

So y = 8.966


For a more detailed method using any single Log base you like:

Take logs of both sides

Log (2^y) = Log (500)

y Log (2) = Log (500) (using basic Log laws, an index inside a Log = multiplying it by the Log)

y = Log (500)/Log (2) = 8.966

Eitherway, it's straightforward high school maths.

(Perhaps I'm showing my age when I say that in the days before calculators, we all had to use logarithms to solve problems. It's a handy way of changing a curved graph into a straight line when solving more difficult real-life or science problems too.)

Another bad idea. What's happening here? How does logarithms help us on GMAT? Do they provide log tables? Do they design questions keeping logarithmic approaches in mind? It's not the right forum to register past.

Your work in bold is acceptable. You should have concluded that 2^y ≥ 500, and if y is a positive integer, then its least value must be 9, because only 2^9 exceeds 500 barrier.

sanju09 wrote:

Mathsbuddy wrote:I agree that counting doubles would be quick.

Anyway, here's a simple way of solving it algebraically too:

2000 * 2^y = 1000000 simplifies to 2^y = 500

Therefore y = Log (500) using Log base 2

So y = 8.966


For a more detailed method using any single Log base you like:

Take logs of both sides

Log (2^y) = Log (500)

y Log (2) = Log (500) (using basic Log laws, an index inside a Log = multiplying it by the Log)

y = Log (500)/Log (2) = 8.966

Eitherway, it's straightforward high school maths.

(Perhaps I'm showing my age when I say that in the days before calculators, we all had to use logarithms to solve problems. It's a handy way of changing a curved graph into a straight line when solving more difficult real-life or science problems too.)

Another bad idea. What's happening here? How does logarithms help us on GMAT? Do they provide log tables? Do they design questions keeping logarithmic approaches in mind? It's not the right forum to register past.

Your work in bold is acceptable. You should have concluded that 2^y ≥ 500, and if y is a positive integer, then its least value must be 9, because only 2^9 exceeds 500 barrier.

I see your point. However, mbahopeful111 asked for an algebraic method, so I gave it. With regard to your last paragraph, I believe the question asked for an approximate number of years, so the closest answer from the list to 8.966, being 9 would be the answer. I didn't know a calculator was not allowed in a GMat exam. I would just type Log(500) in base 2 and move on to the next question. Without a calculator I would count on as mentioned. Thanks.

mbahopeful111 wrote:This question concerns an exponential growth problem. MGMAT presents a simple chart based solution, but I wonder if the the problem can be answered through algebraic manipulation.

"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?"

Can we take the square root of both sides of the equation 1,000,000 = 2,000^(2t)? If so, why isn't the resulting expression 1,000 = 2,000^t? The correct answer is 9 years.

Thanks!

One approach is to WRITE IT OUT.

1 year --> 4000.
2 years --> 8000.
3 year --> 16000.
4 years --> 32000.
5 years --> 64000.
6 years --> 128000.
7 years --> 256000.
8 years --> 512000.
9 years --->1,024,000.

The correct answer is 9 years.

Formula for exponential growth:

Final amount = (original amount) * (multiplier)^(number of changes).

Here, since the product on the righthand side must be equal to or GREATER THAN a target value -- 1,000,000 -- we should use the following:
(original amount) * (multiplier)^(number of changes) ≥ target value.

In the problem at hand:
Original amount = 2000.
Multiplier = 2. (Since the population keeps doubling.)
Number of changes = x, where x is a positive integer.
Target value = 1,000,000.

Plugging these values into the revised formula, we get:
2000 * 2^x ≥ 1,000,000
2^x ≥ 500.
Since 2� = 256 and 2� = 512, x=9.

Thus, the number of years needed = 9.

As 2000 * 2^y = 1000000 simplifies to 2^y = 500

we just need to count how many times to double two, including 2 itself:

2, 4, 8, 16, 32, 64, 128, 256, 512 = 500 approx.

Answer (by counting the number of terms) = 9

Thanks to everyone for your replies. Your explanations really helped me understand the solution to this problem. I now see where I went wrong in my algebraic translation. I also see how we can estimate 9 as the correct answer without an actual logarithmic calculation. Thanks!