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The other rule you can know is that when you add two numbers that are the same with the same exponents, the sum = that same number with exponent +1....see below:

\(2^2 + 2^2 = 2^3\) \(4 + 4 = 8\)

So The first 2 + 2 can be viewed as \(2^1 + 2^1 = 2^2\), then you add that to another \(2^2\) to get \(2^3\) and so on, until you get to \(2^8 + 2^8 = 2^9\)

If I see a question involving exponents and adding, subtracting, multiplying or dividing their bases, etc. i will often try it with something I know the value of easily. Like a base of 2, 3 or 4. Then I take the value of the exponent, do the operation and see if I recognize the result.

Like 2^2 + 2^2 = 8, which is 2^3. So that makes me realize the pattern. \(2^x + 2^x = 2^{x+1}\)

PLEASE NOTE:: This formula really only works for base of 2. If you have base of 3, then you need \(n^x + n^x + n^x = n^{x+1}\) (See how there are 3 n's? Whatever the base is, you need that number of \(n^x\)'s. _________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Although this approach may not suit everyone, it is easy to realise that the question is Geometric Progression, and the formula for a sum of a GP can be used to get the answer. The answer is 2^9

The other rule you can know is that when you add two numbers that are the same with the same exponents, the sum = that same number with exponent +1....see below:

\(2^2 + 2^2 = 2^3\) \(4 + 4 = 8\)

So The first 2 + 2 can be viewed as \(2^1 + 2^1 = 2^2\), then you add that to another \(2^2\) to get \(2^3\) and so on, until you get to \(2^8 + 2^8 = 2^9\)

If I see a question involving exponents and adding, subtracting, multiplying or dividing their bases, etc. i will often try it with something I know the value of easily. Like a base of 2, 3 or 4. Then I take the value of the exponent, do the operation and see if I recognize the result.

Like 2^2 + 2^2 = 8, which is 2^3. So that makes me realize the pattern. \(n^x + n^x = n^{x+1}\)

The formula \(n^x + n^x = n^{x+1}\) does not work for all numbers though. For example \(3^2 + 3^2\) does not equal \(3^3\). Does this formula only apply to a base of 2? _________________

Factorials were someone's attempt to make math look exciting!!!

I just realized that this morning when working on another problem. It works if the base is 2. Essentially, you have to have the same number of numbers with exponents that are the same in order to combine them all with the same base and increase the exponent by 1.

I made the mistake of taking something that works for 2 and applying it to others. That doesn't work as often as I wish it did

brokerbevo wrote:

jallenmorris wrote:

The other rule you can know is that when you add two numbers that are the same with the same exponents, the sum = that same number with exponent +1....see below:

\(2^2 + 2^2 = 2^3\) \(4 + 4 = 8\)

So The first 2 + 2 can be viewed as \(2^1 + 2^1 = 2^2\), then you add that to another \(2^2\) to get \(2^3\) and so on, until you get to \(2^8 + 2^8 = 2^9\)

If I see a question involving exponents and adding, subtracting, multiplying or dividing their bases, etc. i will often try it with something I know the value of easily. Like a base of 2, 3 or 4. Then I take the value of the exponent, do the operation and see if I recognize the result.

Like 2^2 + 2^2 = 8, which is 2^3. So that makes me realize the pattern. \(n^x + n^x = n^{x+1}\)

The formula \(n^x + n^x = n^{x+1}\) does not work for all numbers though. For example \(3^2 + 3^2\) does not equal \(3^3\). Does this formula only apply to a base of 2?

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.