Hey melguy,

When we factor, we need to factor out the term with the smallest exponent.

EXAMPLES:
k^5 + k^3 = k^3(k^2 + 1) [here 3 is the smallest exponent]

m^19 + m^15 = m^15(m^4 + 1) [here 15 is the smallest exponent]

x^(-6) + x^(-4) + x^(-3) = x^(-6)(1 + x^2 + x^3) [here -6 is the smallest exponent]
Aside: -6 is the smallest exponent because -6 < -4 < -3

A few more:
w^x + x^(x+5) = w^x(1 + w^5)

2^x - 2^(x-2) = 2^(x-2)[2^2 - 1]

Does that help?

Cheers,
Brent

melguy wrote:Hi Manpreet and Brent

Thanks for your response. Here is where my confusion lies. We take the smallest value. So 3^-13 is smaller than 3^-11. So we should take 3^-13 as common and not 3^-11. When i solve using 3^-13 i get an incorrect answer but when i 3^-11 as common I get the right answer (O.A).

Let's factor 3^(-11) + 3^(-12) + 3^(-13)
Since the smallest exponent is -13, we get:
3^(-13)[3^2 + 3^1 + 1]
= 3^(-13)[13]

Having said all of that about factoring, we can solve the original question without factoring. Here's how I'd solve the original question:

Original question:


First recognize that asking, "K is how many times Q?" is the same as asking, "What is the value of K/Q?"
For example, asking "15 is how many times 3?" is the same as asking, "What is the value of 15/3?"

So, for the original question, we are asking, "What is the value of [3^(-11)/2 + 3^(-12)/4 + 3^(-13)/6]/3^(-14)?

At this point, I better go to image-mode:


Cheers,
Brent

melguy wrote:Hi Manpreet and Brent

Thanks for your response. Here is where my confusion lies. We take the smallest value. So 3^-13 is smaller than 3^-11. So we should take 3^-13 as common and not 3^-11. When i solve using 3^-13 i get an incorrect answer but when i 3^-11 as common I get the right answer (O.A).

it doesn't matter that 3^-13 is smaller than 3^-11.
When factoring, look at the exponents only.
-13 is smaller than -11, so factor out the 3^-13

Here's another way to look at it:
How would you factor x^3 - x^2?
Since we don't know the value of x, we can't determine whether x^3 is smaller than x^2, or vice versa. So, the relative sizes of x^3 and x^2 don't matter.
All we need to do is recognize that 2 is less than 3, we we factor out x^2 to get x^2(x - 1)

Cheers,
Brent