How To Solve: Exponents

How To Solve: Exponents

Attached pdf of this Article as SPOILER at the top! Happy learning!

Hi All,

I have recently uploaded a video on YouTube to discuss Exponents in Detail:




Following is covered in the video

¤ Simplifying \((-1)^n\)
¤ Simplifying \((-k)^n\)
¤ Simplifying \(0^n\)
¤ Simplifying \(1^n\)
¤ Simplifying \((\frac{x}{y})^a\)
¤ Simplifying \(a^x * a^y\)
¤ Simplifying \(\frac{a^x }{ a^y}\)
¤ Simplifying \(x^a * y^a\)
¤ Simplifying \(x^{a^b}\)
¤ Simplifying \(a^{(-x)}\)
¤ Simplifying \(a^{(\frac{x}{y})}\)
¤ Adding exponents with same base and power
¤ \(x^n – y^n\) is ALWAYS divisible by x-y
¤ \(x^n – y^n\) is divisible by x+y when n is even
¤ \(x^n + y^n\) is divisible by x+y when n is odd
¤ \(x^n + y^n\) is NEVER divisible by x-y

Simplifying \((-1)^n\)

\((-1)^n\)

= - 1 (for all odd values on n)
= + 1 (for all even values of n)

Simplifying \((-k)^n\)

\((-k)^n\)

= 1 if n is 0
= +ve if n is even (except n = 0)
= -ve if n is odd
= 0 if k=0 and n≠0
= not defined if k=0 and n=0

Simplifying \(0^n\)

\(0^n\) = 0 , for all n ≠ 0

Simplifying \(1^n\)

\(1^n\) = 1 ( Always)

Simplifying \((\frac{x}{y})^a\)

\((\frac{𝒙}{𝒚})^𝒂\)= \(\frac{𝒙^𝒂}{𝒚^𝒂}\)

Simplifying \(a^x * a^y\)

If the base of two exponents is same and if we are multiplying the exponents, then we can keep the same base and add the powers.

\(a^x * a^y = a^{( x + y )}\)

Simplifying \(\frac{a^x }{ a^y}\)

If the base of two exponents is same and if we are dividing the exponents, then we can keep the same base and subtract the powers.

\(\frac{𝒂^𝒙}{𝒂^𝒚} = a^{( x – y )}\)

Simplifying \(x^a * y^a\)

If the power of two exponents is same and if we are multiplying the exponents, then we can multiply the bases and keep the same power

\(x^a * y^a = (xy)^a\)

Simplifying \(x^{a^b}\)

\(x^{a^b} = x^{b^a} = x^{ab}\)

Simplifying \(a^{(-x)}\)

\(a^{-x} = \frac{1}{𝑎^𝑥}\)

Simplifying \(a^(\frac{x}{y})\)

\(a^{𝑥/𝑦}= y√(a^x) = (y√a)^x\)

Adding exponents with same base and power

If we are adding two or more exponents with the same power, then we can add them like normal variables

\(x^a + x^a + x^a = 3 * x^a\)

\(x^n – y^n\) is ALWAYS divisible by x-y

Ex: If we take n = 2 then we have, \(x^n - y^n\) = \(x^2 - y^2\) = ( x - y ) * ( x + y) = divisible by x - y

\(x^n – y^n\) is divisible by x+y when n is EVEN

Ex: If we take n = 1 then we have, \(x^n - y^n\) = \(x^1 - y^1\) = ( x - y ) => NOT divisible by x + y
Ex: If we take n = 2 then we have, \(x^n - y^n\) = \(x^2 - y^2\) = ( x - y ) * ( x + y) => divisible by x + y

\(x^n + y^n\) is divisible by x+y when n is ODD

Ex: If we take n = 2 then we have, \(x^n + y^n\) = \(x^2 + y^2\) => there is NO way in which we can express this as (x+y) * some other integer => NOT divisible by x + y
Ex: If we take n = 3 then we have, \(x^n + y^n\) = \(x^3 + y^3\) = ( x + y ) * ( \(x^2 - xy + y^2\)) => divisible by x + y

\(x^n + y^n\) is NEVER divisible by x-y

Ex: If we take n = 2 then we have, \(x^n + y^n\) = \(x^2 + y^2\) => there is NO way in which we can express this as (x-y) * some other integer => NOT divisible by x - y
Ex: If we take n = 3 then we have, \(x^n + y^n\) = \(x^3 + y^3\) = ( x + y ) * ( \(x^2 - xy + y^2\)) => there is NO way in which we can express this as (x-y) * some other integer => NOT divisible by x - y

Hope it helps!
Good Luck!