Last Updated on: 2nd February 2024, 11:30 am

Mathematical Indices

Indices

When a continued product of a variable (axaxa … n times) is expressed in an exponential expression aⁿ, aⁿis called Exponential Expression, ‘a’ is known as Base and ‘n’ is known as Index. The word Indices is plural of Index.

$\displaystyle \mathbf{Ex}.\text{ }{{a}^{2}}=\left( {a\times a} \right),\text{ }{{a}^{3}}=\left( {a\times a\times a} \right),\text{ }{{a}^{m}}=\left( {a\times a\times a~\text{ }\ldots \text{ }m\text{ }times} \right)$

Laws of Indices

Expressions with equal Base

$\displaystyle \bullet \text{ }{{a}^{m}}\times {{a}^{n}}={{a}^{{m+n}}}$
$\displaystyle Ex.\text{ }{{3}^{3}}\times {{2}^{2}}={{3}^{{3+2}}}={{3}^{5}}$

$\displaystyle \bullet \text{ }{{a}^{m}}\div {{a}^{n}}={{a}^{{m-n}}}$
$\displaystyle Ex.\text{ }{{\text{5}}^{4}}\div {{5}^{2}}={{5}^{{4-2}}}={{5}^{2}}$

$\displaystyle \bullet \text{ }{{\left( {{{a}^{m}}} \right)}^{n}}={{a}^{{mn}}}$
$\displaystyle Ex.\text{ }{{\left( {{{\text{2}}^{4}}} \right)}^{3}}={{2}^{{4\times 3}}}={{2}^{{12}}}$

a⁰ = 1 (irrespective of value of a)

Ex: 5⁰ = 1, 7⁰ = 1

The value of any Real number with power zero is always 1. So, whatever be the Base, if the power (Index) is zero, then the value is always 1.
So, 1⁰= 1, 2⁰=1, $\displaystyle {{\left( {\frac{4}{5}} \right)}^{0}}$ =1, $\displaystyle {{\left( {\sqrt{7}} \right)}^{0}}$ =1, $\displaystyle {{\left( {-\sqrt{7}} \right)}^{0}}$ =1
The only exception is Base cannot be zero. So, we cannot write 0⁰=1 (0⁰≠1), This is undefined.

a^p.b^p = (ab)^p
7²x 8²= (7×8)² = 56²
if a^m=aⁿ , where a≠0 and a≠±1 then m=n
So, if the Base is not 1 or -1, the m=n. So, though 1⁵=1 and 1⁷=1, Also, -1⁵= =1 and -1⁷=-1, but we cannot say 5=7. Because this rule is not valid for Base = 1 or -1.
if aⁿ=bⁿ, where n≠0, then a=±b
If value two expressions having same base is equal, then the absolute value of the base must be equal. For example 5² and (-5)² is equal
As explained, this rule is not valid when power is zero,

$\displaystyle \bullet \text{ }{{a}^{{-m}}}=\frac{1}{{{{a}^{m}}}}\text{ and }\frac{1}{{{{a}^{{- m}}}}}={{a}^{m}}$
$\displaystyle Ex.\text{ }{{2}^{{-3}}}=\frac{1}{{{{2}^{3}}}}\text{ and }\frac{1}{{{{2}^{{-5}}}}}={{2}^{5}}$

If a^x = a^y, then x=y
When Base is equal (a) and the Index expression (a^x& a^y) of the equal base (a) is equal (i.e a^x = a^y ), then the index (x & y) are equal (i.e x=y)
Expressions with equal Index
If x^a = y^a, then x=y
When Index is equal (a) and the exponential expression (x^a & y^a) of the equal Index (a) is equal (i.e x^a = y^a ), then the Base (x & y) are equal (i.e x=y)

Indices : Problems

Indices – Problems & Solutions

Ex. 1 : Evaluate and find the value of 2⁴ x 2² x 5².

$\displaystyle {{2}^{4}}\times {{2}^{2}}\times {{5}^{2}}$
$\displaystyle Step\text{ }1~=\text{ }{{2}^{{4+2}}}\times {{5}^{2}}$
$\displaystyle Step\text{ }2~={{2}^{6}}\times {{5}^{2}}=64\times 25=1600$

Ex. 2 : Evaluate 6ab²c³ ´ 4b⁻²c⁻³d.

$\displaystyle 6a{{b}^{2}}{{c}^{3}}\times 4{{b}^{{-2}}}{{c}^{{-3}}}d$
$\displaystyle Step\text{ }1~=\left( {6\times 4} \right)\times a\times {{b}^{2}}\times {{b}^{{-2}}}\times {{c}^{3}}\times {{c}^{{-3}}}\times d$
$\displaystyle \begin{array}{*{20}{l}} {Step\text{ }2~=24a\times {{b}^{{2+(-2)}}}\times {{c}^{{3+(-3)}}}\times d} \ {Step\text{ }3~=24a\times {{b}^{{2-2}}}\times {{c}^{{3-3}}}\times d} \end{array}$
Step 4 = 24ab⁰c⁰d [as b⁰= 1, c⁰= 1]
Step 5 =24ad.

Ex 3 -1: Hindi Audio – Explain in Hindi

Ex : Multiply x⁴y³z² and xy⁵z^-1
Solution: x⁴y³z² and xy⁵z^-1
= x⁴.x .y³.y⁵.z².z^-1
= x⁴⁺¹.y³⁺⁵.z^2-1
= x⁵.y⁸.z

Click Here to play a Video on Indices created by a student, explaining the problem, under dVidya Learn and Earn scheme to earn Money.

Find the value of 27^2/3.
Solution: 27^2/3 = (3³)^2/3
= (3)^3×2/3
= 3²
= 9

Ex – 3-3 Solve $\displaystyle \frac{{{{a}^{3}}{{b}^{2}}}}{{{{a}^{2}}{{b}^{4}}}}$
Solution: $\displaystyle \frac{{{{a}^{3}}{{b}^{2}}}}{{{{a}^{2}}{{b}^{4}}}}$
= a^3-2b^2-4
= a¹b^-2
= a b^-2
= $\displaystyle \frac{a}{{{{b}^{2}}}}$

Ex 3-4 If a^x=b^y=c^z and b²=ac, then prove that (1/x) + (1/z) = 2/y
Step 1: Let a^x=b^y=c^z= k.(as given)
Step 2: Then a=k^1/x (because a^x=k, raising both sides to the power of x), In the same way, b=k^1/y, c=k^1/z
Step 3: Again, b²=ac (Given). So, b²= k^2/yso, b²=k^1/x.k^1/z(because b²=ac)
Step 4 : So, b²= k^1/x+1/z=k^2/y (because b²= k^2/yas derived in step 3)
Step 5: Equating the indices of same base, we get (1/x) + (1/z) = 2/y

Click Here to play a Video on Indices created by a student, explaining the problem, under dVidya Learn and Earn scheme to earn Money

Indices : Problem

Indices – Problems & Solutions

$\displaystyle \mathbf{Ex}.\text{ }\mathbf{1}\text{ }:\text{ }\mathbf{Simplify}\text{ }\frac{{4{{x}^{{-2}}}}}{{{{x}^{-}}^{{1/3}}}}$

$\displaystyle We\text{ }have\text{ }\frac{{4{{x}^{{-2}}}}}{{{{x}^{-}}^{{1/3}}}}$

Step 1 = 4[x^{{−2−(−1/3)}}]

Step 2= 4[x^{(−2 +1/3)}]

Step 3= 4x^{− 5/3} [as the index value (-2 + $\frac{1}{3}$ )=- $\frac{5}{3}$ ]

So, the value of the expression is 4x^{− 5/3}

Ex. 2: Simplify (x^1/2y^−1/2)^4/3 ÷ (x²y⁻¹)^−1/3

We have (x^1/2y^−1/2)^4/3 / (x²y⁻¹)^−1/3

Step 1 : = [(x^1/2)^4/3. (y^−1/2)^4/3] / [(x².y⁻¹)^−1/3]

Step 2 : = [x^{1/2 x 4/3}.y^{−1/2 x 4/3}] / [x^{2 x−1/3}.y^{−1 x −1/3}]

Step 3 : = [x^2/3.y^−2/3] / [x^-2/3.y^1/3]

Step 4= x^{2/3 − (−2/3)}.y^{−2/3 − (1/3)}

Step 5 = x^4/3.y^−3/3

$\displaystyle Step\text{ }6\text{ }=\text{ }{{x}^{{4/3}}}.{{y}^{{-1}}},\text{ }or\text{ }it\text{ }may\text{ }be\text{ }expressed~as\text{ }\frac{{{{x}^{{4/3}}}}}{{{{y}^{{-1}}}}},$

Indices : Problem

Indices – Problems & Solutions

Ex. Evaluate (x^ay^−b)³.(x³y²)^−a.

We have (x^ay^−b)³.(x³y²)^−a

Step 1= (x^a)³.(y^−b)³.(x³)^−a.(y²)^−a

Step 2 = x^3a.y^−3b.x^−3a.y^−2a

Step 3 = x^3a+(−3a).y^{−3b+(−2a)}

Step 4 = x⁰.y^−3b−2a

Step 5 = 1.y^−2a−3b

Step 6 = y^−2a−3b= y^-(2a+3b)

$\displaystyle ^{~}or\text{ }it\text{ }may\text{ }also\text{ }written\text{ }as\text{ }\frac{1}{{{{y}^{{2a+3b}}}}}$

Ex. Simply (x ^m−n)^l ´ (x ^n−l)^m ´ (x^l−m)ⁿ.

The expression (x ^m−n)^l ´ (x ^n−l)^m ´ (x ^{l −m})ⁿ

Step 1 = (x ^ml−nl) ´ (x ^nm−lm) ´ (x ^ln−mn)

Step 2 =x ^{ml−nl+nm−lm+ln−mn}

Step 3 =x⁰ = 1.

So, the value of the expression = 1

Indices : Problem

Indices – Problems & Solutions

Ex. Find the value of x if 2^x+1+ 3 × 2^x−3= 152.

The equation is : 2^x+1 + 3 X 2^x−3 = 152

Step 1 : 2^x X 2 + 3 X 2^x X 2⁻³= 152

Step 2 : 2^x [(2 + 3 X 2⁻³)] = 152

$\displaystyle Step3:\text{ }{{2}^{x}}\times \left( {2+\frac{3}{8}} \right)=152$
$\displaystyle Step4:\text{ }{{2}^{x}}\times \left[ {2+\left( {\frac{3}{{23}}} \right)} \right]=152$
$\displaystyle Step5:\text{ }{{2}^{x}}\times \frac{{19}}{8}=152$
$\displaystyle Step6:\text{ }{{2}^{x}}=152\times \left( {\frac{8}{9}} \right)=64=26$
So, 2^x=2^6, or x=6 [As Base are equal, Index are equal]

Ex. Find the value of

$\displaystyle \left[ {{{{\left( {\frac{{{{x}^{b}}}}{{{{x}^{c}}}}} \right)}}^{{\frac{1}{{bc}}}}}} \right]\left[ {{{{\left( {\frac{{{{x}^{a}}}}{{{{x}^{b}}}}} \right)}}^{{\frac{1}{{ab}}}}}} \right]\left[ {{{{\left( {\frac{{{{x}^{c}}}}{{{{x}^{a}}}}} \right)}}^{{\frac{1}{{ac}}}}}} \right]\text{ }+\text{ }25$

Step 1

Step 2

$\displaystyle =\left[ {{{{({{x}^{{b-c}}})}}^{{^{{\frac{1}{{bc}}}}}}}} \right]\left[ {{{{({{x}^{{a-b}}})}}^{{\frac{1}{{ab}}}}}} \right]\left[ {{{{({{x}^{{c-a}}})}}^{{\frac{1}{{ac}}}}}} \right]+\text{ }25$

$\displaystyle =\left[ {{{x}^{{\frac{{\left( {b-c} \right)}}{{bc}}}}}} \right]\left[ {{{x}^{{\frac{{\left( {a-b} \right)}}{{ab}}}}}} \right]\left[ {{{x}^{{\frac{{\left( {c-a} \right)}}{{ac}}}}}} \right]+25$

Step 3:

$\displaystyle =\text{ }{{x}^{{\left[ {\frac{{{{{^{(}}}^{{ab-ac+ac-bc+bc-ab)}}}}}{{\left( {abc} \right)}}} \right]}}}+\text{ }25=\text{ }{{x}^{0}}+\text{ }25\text{ }=\text{ }1+25\text{ }=\text{ }26$

Indices : Problem

Indices – Problems & Solutions

Ex. If 8^5/3 X 4^−4/5= 2^x, Find the value of x

Solution:

$\displaystyle Step\text{ }1\text{ }LHS=\text{ }{{8}^{{\frac{5}{3}}}}\times {{4}^{{\frac{{-4}}{3}}}}$

$\displaystyle Step\text{ }2\text{ }LHS=\text{ }{{\left( {{{2}^{3}}} \right)}^{{\frac{5}{3}}}}\times {{\left( {{{2}^{2}}} \right)}^{{\frac{{-4}}{5}}}}$

$\displaystyle Step\text{ }3\text{ }LHS={{2}^{5}}\times {{2}^{{\frac{{-8}}{5}}}}$

$\displaystyle Step\text{ }4\text{ }LHS\text{ }=\text{ }{{2}^{{5-\left( {\frac{8}{5}} \right)}}}$

$\displaystyle Step\text{ }5\text{ }LHS\text{ }=\text{ }{{2}^{{\frac{{17}}{5}}}}\text{ }\left[ {as\text{ }5-\frac{8}{5}=\frac{{17}}{5}} \right]$

So, 2^17/5= 2^x. [In the equality, as Base is same, i.e, 2, the Index must be equal].

$\displaystyle So,\text{ }x\text{ }=\frac{{17}}{5}~$

Ex. Find the Value of (a^{m- n})^m+nX (a^n−l)^n+lX (a^l−m)^l+m

Step 1: (a^{m- n})^m+nX (a^n−l)^n+lX (a^l−m)^l+m

Step 2 : a^(m-n).(m+n) X a^(n-l).(n+l)X a^{(l-m) . (l+m)}

Step 3_:a^{m²- n²}X a^{n²- l²}Xa^{l²- m²}

Step 4 : Here, the sum of the indices = (m²-n²+n²-l²+l²-m²) = 0

Step 5 : So, value of the expression may be written as a⁰

So, (a^{m- n})^m+nX (a^n−l)^n+lX (a^l−m)^l+m = a⁰= 1

Indices : Problems

Indices – Problems & Solutions

Ex. If 2^x−1+ 2^x+1 = 160, find the value of x

2^x−1+ 2^x+1= 160.

Step 1 : We may write 2^x+1as 2^(x-1)+2, which again may be written as 2^x-1X 2^2.

Step 2 : So, we may rewrite the expression LHS= 2^x−1+ 2^x+1= 2^x−1+ (2^x-1X 2^2.).

Step 3 : So. the expression may be written as LHS= 2^x−1(1 + 2²)

Step 4 : So, LHS= 2^x−1(1 + 2²) = 160 (as given)

Step 5 : So, 2^x−1 x (1+4) = 160

Step 6 : So, 2^x−1 x 5 = 160

Step 7 : So, 2^x−1 = $\frac{{160}}{5}~$ =32

Step 8: 2^x−1 = 2⁵[In the equality, as Base is same, i.e, 2, the Index must be equal].

Step 9 : So, x-1 = 5, or x = 5+1, or x=6.

Indices : Problems

Indices – Problems & Solutions

Ex. If 3^a = 5^b = 15^c, Find the value of c

Step 1 : Let 3^a =5^b = 15^c =k . So, k=3^a or $\displaystyle {{k}^{{\frac{1}{a}}}}=\text{ }3$

Step 2 : 5^b =k. So, $\displaystyle 5={{k}^{{\frac{1}{b}}}}$

Step 3 : 15^c =k, or $\displaystyle 15={{k}^{{\frac{1}{c}}}}$

$\displaystyle Step\text{ }4\text{ }:\text{ }{{k}^{{\frac{1}{a}}}}\times {{k}^{{\frac{1}{b}}}}=\text{ }{{k}^{{\left( {\frac{1}{a}+\frac{1}{b}} \right)}}}=\text{ }3\times 5\text{ }=15\text{ }=~{{k}^{{\frac{1}{c}}}}\left( {as\text{ }per\text{ }step\text{ }3} \right)$

Step 5 : So, k^(1/a ⁺ ^1/b) = k ^1/c[In the equality base (k) is same, So, the index value $\displaystyle \left( {\frac{1}{a}+\frac{1}{b}} \right)\text{ }\And \text{ }\left( {\frac{1}{c}} \right)$ must be equal]

Step 6 : So, ( $\frac{1}{a}$ + $\frac{1}{b}$ )= $\frac{1}{c}$ ,

Step 7 : $\frac{{a+b}}{{ab}}$ = $\frac{1}{c}$ , i.e c= $\frac{{ab}}{{a+b}}$

Indices Short Problem and Solution

Ex. Which Is Greater, 3√5 Or 4√3

Step 1: First, we compute the value of 3√5.

3√5=√5X3X3 (3 is taken from outside the square root to inside square root, So, it is taken twice)
=√45 (the value inside the square root computed)

Step 2: Next, we compute the value of 4√3.
4√3 =√3X4X4 (4 is taken from outside the square root to inside square root, So, it is taken twice)
=√48 ((the value inside the square root computed)

Step 3: As 48>45
So, 4√3>3√5

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Ex. If a^x=b^y=c^z and b²=ac, then prove that $\frac{1}{x}$ + $\frac{1}{z}$ = $\frac{2}{y}$

Step 1: Let a^x=b^y=c^z= k.(as given)

Step 2: Then $\displaystyle a={{k}^{{\frac{1}{x}}}}$ (because a^x=k, raising both sides to the power of 1/x), In the same way, $\displaystyle b={{k}^{{\frac{1}{y}}}},\text{ }c={{k}^{{\frac{1}{z}}}}$

Step 3: Again, b²=ac (Given).
$\displaystyle So,\text{ }{{b}^{2}}=\text{ }{{k}^{{\frac{2}{y}}}}so,\text{ }{{b}^{2}}={{k}^{{\frac{1}{x}}}}.{{k}^{{\frac{1}{z}}}}\left( {because\text{ }{{b}^{2}}=ac} \right)$

$\displaystyle \mathbf{So},{{b}^{2}}=\text{ }{{k}^{{^{{\frac{1}{x}+\frac{1}{z}}}}}}={{k}^{{^{{\frac{2}{y}}}}}}\left( {because\text{ }{{b}^{2}}=\text{ }{{k}^{{^{{\frac{2}{y}}}}}}as\text{ }derived\text{ }in\text{ }step\text{ }3} \right)$

Step 5: Equating the indices of same base, we get $\frac{1}{x}$ + $\frac{1}{z}$ = $\frac{2}{y}$

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