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Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Indices

We understand that addition can be expressed through multiplication, such as 4 + 4 + 4 + 4 + 4 equaling 5(4) or 5a when generalized. Similarly, repeated multiplication can be represented using powers; for example, 4 × 4 × 4 × 4 × 4 equals 45, and a × a × a × a × a equals a5.

It may be noticed that in the first case 4 is multiplied 5 times and in the second case ‘a’ is multiplied 5 times. In all such cases a factor which multiplies is called the “base” and the number of times it is multiplied is called the “power” or the “index”. Therefore, “4” and “a” are the bases and “5” is the index for both. Any base raised to the power zero is defined to be 1; i.e. a 0. We also define

Indices Chapter Notes | Quantitative Aptitude for CA Foundation

If n is a positive integer, and ‘a’ is a real number, i.e. n ∈ N and a ∈ R (where N is the set of positive integers and R is the set of real numbers), ‘a’ is used to denote the continued product of n factors each equal to ‘a’ as shown below:
an = a × a × a ………….. to n factors.
Here an is a power of “a“ whose base is “a“ and the index or power is “n“.
For example, in 3 × 3 × 3 × 3 = 34 , 3 is base and 4 is index or power.

Law 1

am ×an = am+n , when m and n are positive integers; by the above definition, 
am = a × a ………….. to m factors and an = a × a ………….. to n factors.

∴am × a= (a × a ………….. to m factors) × (a × a ……….. to n factors)

= a × a ………….. to (m + n) factors

= am+n

Now, we extend this logic to negative integers and fractions. First let us consider this for negative integer, that is m will be replaced by –n. By the definition of am × an = am+n ,
we get a–n × an = a–n+n = a= 1
For example 3× 3 5 = (3 × 3 × 3 × 3) × (3 × 3 × 3 × 3 × 3) = 34 + 5 = 39
Again, 3–5 = 1/35 = 1/(3 × 3 × 3 × 3 × 3) = 1/243

Example 1: Simplify 2x1/2 3x-1 if x = 4

Solution: We have 2x 1/2 3x -1

Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Example 2: Simplify 6ab2c× 4b–2c–3d.

Solution: 6ab2c3 × 4b–2c–3d

= 24 × a × b2 × b–2 × c3 × c–3 × d

= 24 × a × b2+(–2) × c3+(–3) × d

= 24 × a × b2 – 2 × c3 – 3 × d

= 24a b0 × c0 × d

Law 2

am/an = am–n , when m and n are positive integers and m > n.
By definition, am = a × a ………….. to m factors

Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Now we take a numerical value for a and check the validity of this Law

Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Example 3: Find the value of Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Solution:

Indices Chapter Notes | Quantitative Aptitude for CA Foundation

= 4x-1 - (-1/3

= 4x-1 + 1/3

Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Example 4: Simplify Indices Chapter Notes | Quantitative Aptitude for CA Foundation.

Solution:

Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Law 3

(am)n = amn . where m and n are positive integers By definition (am)n 
= a× am × am …….. to n factors 
= (a × a …….. to m factors) a × a ×…….. to m factors……..to n times 
= a × a …….. to mn factors 
= amn

Following above, (am)n = (am)p/q

(We will keep m as it is and replace n by p/q, where p and q are positive integers)

Indices Chapter Notes | Quantitative Aptitude for CA Foundation

If we take the qth root of the above we obtain

Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Now with the help of a numerical value for a let us verify this law.

(24)3 = 24 × 24 × 24

= 24+4+4
= 212
= 4096

Law 4

(ab)n = an.bn when n can take all of the values.
For example 6= (2 × 3) 3 = 2 × 2 × 2 × 3 × 3 × 3 = 23 × 33

First, we look at n when it is a positive integer. Then by the definition, we have
(ab)= ab × ab ……………. to n factors
= (a × a ……..……. to n factors) × (b × b …………. n factors)
= an × bn

When n is a positive fraction, we will replace n by p/q.

Then we will have (ab)n = (ab)p/q

The qth power of (ab)p/q = {(ab)(p/q)}q = (ab)p

Example 5: Simplify (xa.y–b)3 . (x3 y2)a

Solution: (xa.y–b)3 . (x3 y2)a

Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Example 6: Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Solution: Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Example 7: Find x, if Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Solution: Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Indices Chapter Notes | Quantitative Aptitude for CA Foundation

[If base is equal, then power is also equal]

Indices Chapter Notes | Quantitative Aptitude for CA Foundation

∴ x = 1

Example 8: Find the value of k from Indices Chapter Notes | Quantitative Aptitude for CA Foundation

Solution: Indices Chapter Notes | Quantitative Aptitude for CA Foundation 
or, (32 × 1/2) –7 × (3½)–5 = 3
or, 3-7-5/2  = 3
or, 3 –19/2 = 3k
or, k = –19/2

The document Indices Chapter Notes | Quantitative Aptitude for CA Foundation is a part of the CA Foundation Course Quantitative Aptitude for CA Foundation.
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FAQs on Indices Chapter Notes - Quantitative Aptitude for CA Foundation

1. What are the fundamental laws of logarithms?
Ans. The fundamental laws of logarithms include: 1. <b>Product Law:</b> \( \log_b(m \cdot n) = \log_b(m) + \log_b(n) \) 2. <b>Quotient Law:</b> \( \log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n) \) 3. <b>Power Law:</b> \( \log_b(m^n) = n \cdot \log_b(m) \) These laws help simplify complex logarithmic expressions.
2. How do you change the base of a logarithm?
Ans. To change the base of a logarithm, you can use the change of base formula: \[ \log_b(a) = \frac{\log_k(a)}{\log_k(b)} \] where \( k \) is any positive number different from 1. This allows you to convert logarithms to a more convenient base.
3. What is the significance of logarithms in solving exponential equations?
Ans. Logarithms are significant in solving exponential equations because they allow you to bring the exponent down as a coefficient, making it easier to isolate the variable. For example, if you have \( b^x = y \), you can take the logarithm of both sides to get \( x = \log_b(y) \).
4. Can logarithms be used with any base?
Ans. Yes, logarithms can be used with any positive base \( b \) where \( b \neq 1 \). Common bases include base 10 (common logarithm) and base \( e \) (natural logarithm). The choice of base can simplify calculations depending on the context.
5. What are some real-world applications of logarithms?
Ans. Logarithms have several real-world applications, including: 1. <b>Sound Intensity:</b> The decibel scale, which measures sound intensity, is logarithmic. 2. <b>Earthquake Measurement:</b> The Richter scale, which measures earthquake magnitude, is based on logarithmic values. 3. <b>pH Scale:</b> The pH scale used in chemistry to measure acidity is logarithmic. These applications highlight the importance of logarithmic functions in various fields.
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