**Indices**

**Indices**

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

**Laws of Indices**

**Expressions with equal Base**

^{ }

- a
^{0}= 1 (irrespective of value of a)

Ex**: **5^{0} = 1, 7^{0} = 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,

- 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 2^{4} x 2^{2} x 5^{2}.^{ }

**Ex. 2 : Evaluate 6ab ^{2}c^{3} **

**´**

**4b**

^{−2}c^{−3}d.

Step 4 = 24ab^{0}c^{0}d [as b^{0 }= 1, c^{0 }= 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**

Step 1 = 4[x^{{−2−(−1/3)}}]

Step 2= 4[x^{(−2 +1/3)}]

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

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

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

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

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

**Indices : Problem**

**Indices – Problems & Solutions**

**Ex. Evaluate** (x^{a }y^{−b})^{3}.(x^{3 }y^{2})^{−a}.

We have (x^{a }y^{−b})^{3}.(x^{3 }y^{2})^{−a}

^{ }Step 1= (x^{a})^{3}.(y^{−b})^{3}.(x^{3})^{−a}.(y^{2})^{−a}

Step 2 = x^{3a}.y^{−3b}.x^{−3a}.y^{−2a}

Step 3 = x^{3a+(−3a) }.y^{−3b+(−2a)}

Step 4 = x^{0}.y^{−3b−2a}

Step 5 = 1.y^{−2a−3b}

Step 6 = y^{−2a−3b }= 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 =x^{0} = 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

So, 2^{x}=2^{6, } or x=6 [As Base are equal, Index are equal]

**Ex. Find the value of**

**Step 1**

**Step** **2**

**Step 3:**

**Indices : Problem**

**Indices – Problems & Solutions**

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

**Solution:**

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

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

Step 1: (a^{m- n})^{m+n }X (a^{n−l})^{n+l }X (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 }X_{ }a^{l²- m²}

Step 4 : Here, the sum of the indices = (m^{2}-n^{2}+n^{2}-l^{2}+l^{2}-m^{2}) = 0

Step 5 : So, value of the expression may be written as a^{0}

So, (a^{m- n})^{m+n }X (a^{n−l})^{n+l }X (a^{l−m})^{l+m} = a^{0 }= 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+1 }as 2^{(x-1)+2}, which again may be written as 2^{x-1}X 2^{2.}

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

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

Step 4 : So, LHS= 2^{x−1}(1 + 2^{2}) = 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} = =32

Step 8: 2^{x−1} = 2^{5 }[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

Step 2 : 5^{b} =k. So,

Step 3 : 15^{c} =k, or

Step 5 : So, k^{(1/a} ^{+} ^{1/b)} = k ^{1/c }[In the equality base (k) is same, So, the index value must be equal]

Step 6 : So, ( + )=,

Step 7 : = , i.e c=

**Ex. If a ^{x}=b^{y}=c^{z} and b^{2}=ac, then prove that **

**+**

**=**

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

**Step 2**: Then (because a^{x}=k, raising both sides to the power of 1/x), In the same way,

**Step 3: **Again, b^{2}=ac (Given).

**Step 5: **Equating the indices of same base, we get + =

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