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=)0 =1, 0=1, 0=1
The only exception is Base cannot be zero. So, we cannot write 00=1 (00≠1), This is undefined.

ap.bp = (ab)p
72 x 82 = (7×8)2 = 562

if am=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.z2-1
= x5.y8.z

Click Here to play a Video on Indices

Ex 3-2 Find the value of 272/3.

Solution: 272/3 = (33)2/3
= (3)3×2/3
= 32
= 9

Ex – 3-3 Solve a3b2/a2b4

Solution: a3b2/a2b4
= a3-2b2-4
= a1b-2
= a b-2
=a/b2

Ex 3-4 If ax=by=cz and b2=ac, then prove that (1/x) + (1/z) = 2/y

Atep 1: Let ax=by=cz = k.(as given)
Step 2: Then a=k1/x (because ax=k, raising both sides to the power of x), In the same way, b=k1/y, c=k1/z
Step 3: Again, b2=ac (Given). So, b2= k2/y so, b2=k1/x.k1/z (because b2=ac)
Step 4 : So, b2= k1/x+1/z =k2/y (because b2= k2/y as 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

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= $\frac{{ab}}{{a+b}}$

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 $\frac{1}{x}$ + $\frac{1}{z}$ = $\frac{2}{y}$

To see PDF

Indices

Indices : Problems

Indices : Problem

Indices : Problem

Indices : Problem

Indices : Problem

Indices : Problems

Indices : Problems

Like this:

Ask Question, Get Answer