SOLVING EXPONENTIAL AND LOGARITHMIC INEQUALITIES

Problem 1 :

Simplify 

Solve 3^x < 20

Solution :

3^x < 20

x < log₃ 20

Using the properties of logarithm, we get

x < log 20 / log 3

x < 1.3010 / 0.4771

x < 2.72

Problem 2 :

Simplify

log x ≤ 2

Solution :

log x ≤ 2

x ≤ 10²

x ≤ 100

Because log x is only defined when x > 0, the solution is 0 < x ≤ 100.

Problem 3 :

Solve the inequality.

a) e^x < 2

b) 10^{2x - 6} > 3

c) log x + 9 <  45

d) 2 ln x − 1 > 4

Solution :

a) e^x < 2

x < ln 2

x < 0.69

b) 10^{2x - 6} > 3

2x - 6 > log 3 

2x - 6 > 0.477

2x > 0.477 + 6

2x > 6.477

x > 6.477/2

x > 3.2385

So, the possible solution is 3.2385 < x < ∞.

c) log x + 9 <  45

log x < 45 - 9

log x < 36

x < 10³⁶

So, the solution is 0 < x < 10³⁶

d) 2 ln x − 1 > 4

2 ln x > 4 + 1

2 ln x > 5

ln x > 5/2

ln x > 2.5

x > e^2.5

x > 12.18

So, the solution is 12.18 < x < ∞

Problem 4 :

What is the solution to the logarithmic inequality −4 log₂ x ≥ −20?

a) x ≤ 32      b) 0 ≤ x ≤ 32     c)  0 < x ≤ 32     d) x ≥ 32

Solution :

−4 log₂ x ≥ −20

Dividing by -4, we get 

log₂ x ≤ 5

x ≤ 2⁵

x ≤ 32

Do remember : log 1 = 0, log 0 = undefined

So, 0 is not possible value of x, 0 < x ≤ 32. then option c is correct.

Problem 5 :

Solve 9^x > 54

Solution :

9^x > 54

x > log₉ 54

Using the properties of logarithms,

x > log 54 / log 9

x > 1.732/0.954

x > 1.81

So, the possible solution is 1.81 < x < ∞.

Problem 6 :

Solve 4^x ≤ 36

Solution :

4^x ≤ 36

x ≤ log₄ 36

Using the properties of logarithms,

x ≤ log 36 / log 4

x ≤ 1.556 / 0.602

x ≤ 2.58

So, the possible solution is 0 < x ≤ 2.58

Problem 7 :

Solve ln x ≥ 3

Solution :

ln x ≥ 3

x ≥ e³

x ≥ 20.08

So, the possible solution is 20.08 ≤ x < ∞.

Problem 8 :

Solve log₄ x < 4

Solution :

log₄ x < 4

x < 4⁴

x < 256

So, the solution is 0 < x < 256.

Problem 9 :

log_√3 (x + 1) - log_√3 (x - 1) < log₃ 4

Solution :

log_√3 (x + 1) - log_√3 (x - 1) < log₃ 4

Using properties of logarithm,

log m - log n = log (m/n)

log_√3 [(x + 1)/(x - 1)] < log₃ 4

1/log _{[(x + 1)/(x - 1)]}√3 < log₃ 4

1/log _{[(x + 1)/(x - 1)]}(3)^1/2 < log₃ 4

2/log _{[(x + 1)/(x - 1)]}(3) < log₃ 4

2 log₃[(x + 1)/(x - 1)] < log₃ 4

Dividing by 2 on both sides, we get

log₃[(x + 1)/(x - 1)] < (log₃ 4) / 2

log₃[(x + 1)/(x - 1)] < (1/2) log₃ 4

log₃[(x + 1)/(x - 1)] < log₃ √4

log₃[(x + 1)/(x - 1)] < log₃ 2

(x + 1)/(x - 1) < 2

Subtracting 2 on both sides, we get

[(x + 1) / (x - 1)] - 2 < 0

[x + 1 - 2(x - 1)] / (x - 1) < 0

(x + 1 - 2x + 2) / (x - 1) < 0

(-x + 3) / (x - 1) < 0

-x + 3 < 0

-x < -3

x > 3

x ∈ (1, 3)

Problem 10 :

log₄ (2x² + x + 1) - log₂ (2x - 1) ≤ 1

Solution :

log₄ (2x² + x + 1) - log₂ (2x - 1) ≤ 1

[1/log_(2x2 _{+ x + 1)} 4] - log₂ (2x - 1) ≤ 1

[1/log_(2x2 _{+ x + 1)} 2²] - log₂ (2x - 1) ≤ 1

[(1/2) (1/log_(2x2 _{+ x + 1)} 2)] - log₂ (2x - 1) ≤ 1

[log₂ (2x² + x + 1)] - 2log₂ (2x - 1)]/2 ≤ 1

log₂ (2x² + x + 1)] - 2log₂ (2x - 1) ≤ 2

log₂ (2x² + x + 1) - log₂ (2x - 1)² ≤ 2

log₂ [(2x² + x + 1)/(2x - 1)² ] ≤ 2

Moving logarithm to the other side, we get

[(2x² + x + 1)/(2x - 1)² ] ≤ 4

[(2x² + x + 1)/(2x - 1)² ] - 4 ≤ 0

[(2x² + x + 1) - 4(2x - 1)² ]/(2x - 1)²   ≤ 0

(2x² + x + 1) - 4[(2x)² - 2(2x)(1) + 1² ] ≤ 0

(2x² + x + 1) - 4[4x² - 4x + 1] ≤ 0

2x² + x + 1 - 16x² + 16x - 4 ≤ 0

- 14x² + 17x - 3 ≤ 0

14x² - 17x + 3 ≥ 0

14x² - 17x + 3 ≥ 0

(x - 1)(14x - 3) ≥ 0

x ≥ 0 and x ≥ 3/14

Finding domain of the logarithmic functions :

The value of x ∈ (1, ∞)