Exponential and logarithmic equations worksheet – Free JEE Math Tutorial

Exponential and Logarithmic Equations

Key Definitions

  1. Exponential Equation: An equation where the unknown variable appears in the exponent of a constant base, typically of the form af(x) = b.
  2. Logarithmic Equation: An equation involving the logarithm of an expression containing the variable, such as loga(f(x)) = b.
  3. Logarithmic Domain: For any expression loga(g(x)), the domain is strictly defined by: g(x) > 0, a > 0, and a ≠ 1.
  4. Base Change Formula: A fundamental identity for solving equations with different bases: loga b = (logc b) / (logc a).

Theory and Concepts

1. Solving Exponential Equations

Exponential equations are generally solved using two primary methods:

  • Same Base Method: If af(x) = ag(x) and a > 0, a ≠ 1, then f(x) = g(x).
  • Substitution Method: Equations like a · (kx)² + b · kx + c = 0 are reduced to a quadratic form by letting t = kx (where t > 0).

2. Solving Logarithmic Equations

The equation loga f(x) = loga g(x) implies f(x) = g(x), provided f(x) > 0 and g(x) > 0.

Warning: Always check the final solutions against the original domain constraints to identify extraneous roots.

3. Transformation to Polynomial Form

A common technique for complex exponential equations is taking the natural logarithm on both sides:

y = [f(x)]g(x) ⇒ ln y = g(x) ln f(x)

Figure 15.1: Inverse relationship between Exponential and Logarithmic functions.

Solved Examples

Solved Example 1.1

Problem: Solve for x: 4x – 3 · 2x+1 + 8 = 0.

Solution

Rewrite: (2x)² – 6(2x) + 8 = 0.
Let 2x = t (t > 0).
t² – 6t + 8 = 0 ⇒ (t-2)(t-4) = 0.
t = 2 ⇒ 2x = 2¹ ⇒ x = 1.
t = 4 ⇒ 2x = 2² ⇒ x = 2.
Roots: {1, 2}.

Solved Example 1.2

Problem: Solve log2(x-1) + log2(x-3) = 3.

Solution

Domain: x-1 > 0 and x-3 > 0 ⇒ x > 3.
log2[(x-1)(x-3)] = 3 ⇒ (x-1)(x-3) = 2³.
x² – 4x + 3 = 8 ⇒ x² – 4x – 5 = 0.
(x-5)(x+1) = 0 ⇒ x = 5 or x = -1.
Check domain: x=5 is valid; x=-1 is extraneous.
Answer: x = 5.

Solved Example 1.3

Problem: Solve xlogx(x+3) = 7.

Solution

Using the identity aloga n = n:
x + 3 = 7 ⇒ x = 4.
Check constraints: x > 0, x ≠ 1. x=4 is valid. Answer: x = 4.

Solved Example 1.4

Problem: Solve 3x+1 = 5x-1.

Solution

(x+1) log 3 = (x-1) log 5
x log 3 + log 3 = x log 5 – log 5
x(log 5 – log 3) = log 5 + log 3
x = log(15) / log(5/3).

Solved Example 1.5

Problem: Solve for x: logx 2 + log2 x = 2.5.

Solution

Let log2 x = t. Then 1/t + t = 2.5.
2t² – 5t + 2 = 0 ⇒ (2t-1)(t-2) = 0.
t = 2 ⇒ x = 2² = 4.
t = 1/2 ⇒ x = 21/2 = √2.
Roots: {4, √2}.

Solved Example 1.6

Problem: Find the number of real solutions of ex = sin x.

Solution

For x > 0, ex > 1 and sin x ≤ 1 (No solution).
For x ≤ 0, ex is a positive decreasing function approaching 0, while sin x oscillates. The curve will intersect the positive cycles of the sine wave infinitely many times as x → -∞.
Answer: Infinite solutions.

Solved Example 1.7

Problem: Solve log3(5 + 4 · 3x-1) = 2x.

Solution

5 + 4 · 3x-1 = 32x. Let 3x = t.
5 + 4t/3 = t² ⇒ 3t² – 4t – 15 = 0.
(3t + 5)(t – 3) = 0.
t = 3 ⇒ 3x = 3 ⇒ x = 1.

Solved Example 1.8

Problem: If log10 2, log10(2x-1), log10(2x+3) are in A.P., find x.

Solution

2 log(2x-1) = log 2 + log(2x+3)
(2x-1)² = 2(2x+3). Let 2x = y.
y² – 2y + 1 = 2y + 6 ⇒ y² – 4y – 5 = 0.
(y-5)(y+1) = 0 ⇒ y = 5 (since 2x > 0).
2x = 5 ⇒ x = log2 5.

Solved Example 1.10

Problem: Solve | log2 x | = 2 – x.

Solution

Case 1: x ≥ 1. log2 x = 2 – x. By observation, x = 2 is a root.
Case 2: 0 < x < 1. -log2 x = 2 – x ⇒ log2 x = x – 2. Graphical analysis shows one intersection in (0,1).
Total: 2 solutions.

Concept Application Exercise 1.1

  1. Solve for x: 9x – 4 · 3x + 3 = 0.

  2. Solve log(x² – 1) – log(x – 1) = 0.

  3. Solve 2x+2 + 2x-1 = 9.

  4. Find the value of x satisfying xlog10 x = 100x.

  5. Solve logx 3 + log3 x = log√x 3 + log3 √x + 0.5.

  6. Find the number of real roots of 2x + 3x + 4x = 3.

  7. Solve logx+3(x² – x) = 1.

  8. If ax = by = cz = dw and a, b, c, d are in G.P., prove x, y, z, w are in H.P.

  9. Solve 51+log5 cos x = 2.5.

  10. Solve the system: 3x · 2y = 576 and log√2(y-x) = 4.