Help with doing exponents
Introduction .
A number of the form ax is called the exponent form. This is a very convenient way of expressing a set of numbers being multiplied for so many times. Suppose, if I want to multiply 3 five times, it is easy to write as follows: 3 `xx` 3 `xx` 3` xx` 3 `xx` 3 = 243.But if the same thing if we want to multiply 1000 times, how long we can write them to get the answer. But, exponent give us a convenient way of representing then as `3^1000` . Now we have come to understand the reason behind having the exponent form. Now if we need to handle more than one value of the exponent, we need to learn some convenient method to evaluate them. For that purpose, we need to do know the following formula's. They are as follows:
(i) am `xx` an = am + n [ Here , if we need to multiply two exponents, result is obtained by adding the powers of the exponents]
(ii) `a^m / a^n` = am – n [ Here , if we need divide two exponents, result is obtained by subtracting the numerator powers with the denominator power of the exponents]
(iii) (am) n = amn [ When we take power over the power of an exponent, the result is obtained by multiplying the powers]
(iv) (a b) m = am `xx` bm
(v) (`a/b`)m = `a^m / b^m`
(vi) a – m = `1 / a^m` [ This is applicable for both the ways, that is a m = `1 / a^[-m]` ]
(vii) a0 = 1.
By applying these above formula’s we can evaluate the exponents in a more comfortable manner.
Now let us do few problems of this kind.
Example problems helpful with doing exponents:
Ex 1: Evaluate: (i) `9^(3/2)` , (ii) `8^(4/3)` , (iii) `(- 32) ^(3/5)` .
Solution: (i) 9^(3/2) = (3^2)^(3/2) = 33 = 27.
(ii) `8 ^(4/3)` = `(2^3)^(4/3)` = 24 = 16.
(iii) (-32)^( 3/5) = `((-2)^5) ^(3/5)` = (-2)3 = - 8.
Ex 2: Simplify: x ^(4/3) ÷ x^(-2/3).
Solution:` x ^(4/3) / x ^(-2/3)` = `x ^[(4/3) + 2/3]` = `x ^[(4+2) / 3]` .
= `x ^(6/3)` = x2.
Ex 3: Evaluate: 53 `xx` `25^ (-3/2)`
= 53 `xx` `(5^2) (-3/2)` = 53 `xx` 5-3 = `5^(3 -3)` = 50 = 1
Ex 4: Evaluate: `( 27 /125 )^(2/3) xx ( 9/ 25)^(-3/2)`
Solution: `( 27 /125 )^(2/3) xx ( 9/ 25)^(-3/2)` = `( 3^3 /5^3 )^(2/3) xx ( 3^2 / 5^2)^(-3/2)`
= `(( 3/5)^3 )^(2/3) xx ((3/5)^2)^(-3/2)` .
= `(3/5)^2 xx (3/5)^(-3) `
= `(3/5)^ (2-3)`
= `(3/5)^(-1)` = `5/3` .
Ex 5: Evaluate: `(16/81)^[-3/4] xx (49/9)^[3/2] -: (343/216)^[2/3]` .
Solution:` ( 2^4 / 3^4) ^[-3/4] xx ( 7^2 / 3^2)^(3/2) -: ( 7^3 /6^3)^[2/3]` .
= `((2/3)^4)^[ -3/4] xx ((7/3)^2)^[3/2] -: ((7/6)^3)^[2/3]`
= `(2/3)^[ -3] xx (7/3)^3 -: (7/6)2`
= `[2^-3] /3^[ -3] xx 7^3 / 3^3 xx 6^2 / 7^2` = `(2^[-3] xx 7^[3 - 2] xx 6^2) / (3^[-3 + 3])`
= `2^[-3] xx 7 xx (2 xx3)^2 / 3^0` = `7 xx 2^2 xx 3^2 / 2^3` .
= `[7 xx 9] / [2^ [3 ** 2]]` = `63/2` = 31.5.
Practice problems on exponents:
1. Evaluate: `[3^[x+2] ** 3^[x+1]]/[4 xx 3^x ** 3^x]` .
[Ans : 2]
2. Evaluate:` [5^[x+10] xx 25^[3x **4]]/[5^[7x]]` .
[Ans: 25]
A number of the form ax is called the exponent form. This is a very convenient way of expressing a set of numbers being multiplied for so many times. Suppose, if I want to multiply 3 five times, it is easy to write as follows: 3 `xx` 3 `xx` 3` xx` 3 `xx` 3 = 243.But if the same thing if we want to multiply 1000 times, how long we can write them to get the answer. But, exponent give us a convenient way of representing then as `3^1000` . Now we have come to understand the reason behind having the exponent form. Now if we need to handle more than one value of the exponent, we need to learn some convenient method to evaluate them. For that purpose, we need to do know the following formula's. They are as follows:
(i) am `xx` an = am + n [ Here , if we need to multiply two exponents, result is obtained by adding the powers of the exponents]
(ii) `a^m / a^n` = am – n [ Here , if we need divide two exponents, result is obtained by subtracting the numerator powers with the denominator power of the exponents]
(iii) (am) n = amn [ When we take power over the power of an exponent, the result is obtained by multiplying the powers]
(iv) (a b) m = am `xx` bm
(v) (`a/b`)m = `a^m / b^m`
(vi) a – m = `1 / a^m` [ This is applicable for both the ways, that is a m = `1 / a^[-m]` ]
(vii) a0 = 1.
By applying these above formula’s we can evaluate the exponents in a more comfortable manner.
Now let us do few problems of this kind.
Example problems helpful with doing exponents:
Ex 1: Evaluate: (i) `9^(3/2)` , (ii) `8^(4/3)` , (iii) `(- 32) ^(3/5)` .
Solution: (i) 9^(3/2) = (3^2)^(3/2) = 33 = 27.
(ii) `8 ^(4/3)` = `(2^3)^(4/3)` = 24 = 16.
(iii) (-32)^( 3/5) = `((-2)^5) ^(3/5)` = (-2)3 = - 8.
Ex 2: Simplify: x ^(4/3) ÷ x^(-2/3).
Solution:` x ^(4/3) / x ^(-2/3)` = `x ^[(4/3) + 2/3]` = `x ^[(4+2) / 3]` .
= `x ^(6/3)` = x2.
Ex 3: Evaluate: 53 `xx` `25^ (-3/2)`
= 53 `xx` `(5^2) (-3/2)` = 53 `xx` 5-3 = `5^(3 -3)` = 50 = 1
Ex 4: Evaluate: `( 27 /125 )^(2/3) xx ( 9/ 25)^(-3/2)`
Solution: `( 27 /125 )^(2/3) xx ( 9/ 25)^(-3/2)` = `( 3^3 /5^3 )^(2/3) xx ( 3^2 / 5^2)^(-3/2)`
= `(( 3/5)^3 )^(2/3) xx ((3/5)^2)^(-3/2)` .
= `(3/5)^2 xx (3/5)^(-3) `
= `(3/5)^ (2-3)`
= `(3/5)^(-1)` = `5/3` .
Ex 5: Evaluate: `(16/81)^[-3/4] xx (49/9)^[3/2] -: (343/216)^[2/3]` .
Solution:` ( 2^4 / 3^4) ^[-3/4] xx ( 7^2 / 3^2)^(3/2) -: ( 7^3 /6^3)^[2/3]` .
= `((2/3)^4)^[ -3/4] xx ((7/3)^2)^[3/2] -: ((7/6)^3)^[2/3]`
= `(2/3)^[ -3] xx (7/3)^3 -: (7/6)2`
= `[2^-3] /3^[ -3] xx 7^3 / 3^3 xx 6^2 / 7^2` = `(2^[-3] xx 7^[3 - 2] xx 6^2) / (3^[-3 + 3])`
= `2^[-3] xx 7 xx (2 xx3)^2 / 3^0` = `7 xx 2^2 xx 3^2 / 2^3` .
= `[7 xx 9] / [2^ [3 ** 2]]` = `63/2` = 31.5.
Practice problems on exponents:
1. Evaluate: `[3^[x+2] ** 3^[x+1]]/[4 xx 3^x ** 3^x]` .
[Ans : 2]
2. Evaluate:` [5^[x+10] xx 25^[3x **4]]/[5^[7x]]` .
[Ans: 25]