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Grade 10 Notes of Indices || Compulsory Mathematics

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Exponential equations are solved by making the bases equal on both sides; if bases are same, indices are equal (2^x = 8 → x = 3). When bases are different, express everything in terms of one common power (let 2^x = a) and form a simple or quadratic equation. Common types include direct form (4^x = 64), factored form (3^x + 3^x = 6 × 3^x), and substitution (2^x + 1/2^x = 5/2 → let a = 2^x). Complex equations become quadratic in a (like a² – 10a + 9 = 0) and give two values of x.

1. Definition

An exponential equation is an equation where the variable appears in the exponent.

General form:
a^(f(x)) = b, where a > 0, a ≠ 1

Key points:

If bases are equal, exponents are equal.

Often appears in growth/decay, compound interest, population problems.

2. Laws of Indices

LawFormulaExample
Multiplicationa^m × a^n = a^(m+n)2^3 × 2^4 = 2^7 = 128
Divisiona^m ÷ a^n = a^(m−n)5^6 ÷ 5^2 = 5^4 = 625
Power of a power(a^m)^n = a^(m×n)(3^2)^3 = 3^6 = 729
Zero exponenta^0 = 17^0 = 1
Negative exponenta^(−n) = 1 / a^n2^(−3) = 1/8
Fractional exponenta^(1/n) = nth root of a8^(1/3) = 2

3. Solving Exponential Equations – Golden Rule

Steps:

Try to write both sides with the same base.

If bases are equal, equate exponents: a^(f(x)) = a^(g(x)) → f(x) = g(x)

Solve the resulting equation (linear or quadratic).

Check solutions in original equation.

4. Types of Exponential Equations

Type 1: Direct Same Base

Bases are already same → equate exponents.

Example:
Q: 2^x = 8
2^x = 2^3
x = 3

Type 2: Change to Same Base

Rewrite both sides with a common base.

Example:
Q: 4^(x−2) = 1/16
(2^2)^(x−2) = 2^(−4)
2^(2x−4) = 2^(−4)
2x − 4 = −4
x = 0

Type 3: Factor Out Common Term

Factor common powers of the base.

Example:
Q: 3^(x+1) + 3^x = 4 × 3^x
3^x (3 + 1) = 4 × 3^x
4 = 4 → True for all x

Type 4: Let Common Term = a (Substitution Method)

Let a = base^x to convert into quadratic form.

Example:
Q: 2^x + 1/2^x = 5/2
Let 2^x = a → a + 1/a = 5/2
2a^2 − 5a + 2 = 0
(2a−1)(a−2) = 0
a = 2 or a = 1/2
2^x = 2 → x = 1
2^x = 1/2 → x = −1

Type 5: Multiple Terms (Quadratic in Disguise)

Treat base^x as a variable, solve quadratic.

Example:
Q: 5 × 4^(x+1) − 16^x = 64
Let 4^x = a → 20a − a^2 = 64
a^2 − 20a + 64 = 0
(a−4)(a−16) = 0
a = 4 → x = 1
a = 16 → x = 2

5. Important Question

Q1: 2^x + 2^(x−1) = 6
(3/2) × 2^x = 6
2^x = 4
x = 2

Q2: 5^(x+1) + 5^x = 150
5^x × 6 = 150
5^x = 25
x = 2

Q3: 4^x + 1/4^x = 17/8
Let 4^x = a
a + 1/a = 17/8
8a^2 − 17a + 8 = 0
a = 1 or a = 2
x = 0 or x = 1

Q4: 3^(2x) − 10 × 3^x + 9 = 0
Let 3^x = a
a^2 − 10a + 9 = 0
(a−1)(a−9) = 0
a = 1 → x = 0
a = 9 → x = 2

Q5: 2^(x+1) + 2^x + 2^(x−1) = 28
2^x × (2 + 1 + 1/2) = 28
2^x × 7/2 = 28
2^x = 8
x = 3

Q6: 4^(2x) − 5 × 4^x + 6 = 0
Let 4^x = a
a^2 − 5a + 6 = 0
(a−2)(a−3) = 0
a = 2 → x = 1/2
a = 3 → x = log4(3)

6. Important Tips

1.If possible, convert both sides to same base.

2.Factor quadratic form carefully.

3.Check all answers in the original equation.

4.Substitute base^x = a for complicated sums like a + 1/a.

5.Negative exponents may lead to fractions.

6.Always write powers as integers when possible to simplify

Visit this link for further practice

https://besidedegree.com/exam/s/academic

 

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