Pooja asked in Science & MathematicsMathematics · 7 months ago

# Find the least value of : 2 log_10 (x) - log_x (0.01) for x > 1?

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• 7 months ago

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• 7 months ago

f(x) = 2.Log[10](x) - Log[x](0.01) → for x > 1

f(x) = 2.Log[10](x) - Log[x](0.01) → recall: Log[a](x) = Ln(x) / Ln(a) ← where a is the base

f(x) = 2.[Ln(x) / Ln(10)] - [Ln(0.01) / Ln(x)] → you know that: 0.01 = 1/100

f(x) = 2.[Ln(x) / Ln(10)] - [Ln(1/100) / Ln(x)] → you know that: Ln(a/b) = Ln(a) - Ln(b)

f(x) = 2.[Ln(x) / Ln(10)] - [{Ln(1) - Ln(100)} / Ln(x)] → you know that: Ln(1) = 0

f(x) = 2.[Ln(x) / Ln(10)] - [- Ln(100) / Ln(x)] → you know that: 100 = 10²

f(x) = 2.[Ln(x) / Ln(10)] - [- Ln(10²) / Ln(x)] → recall: Ln(x^a) = a.Ln(x)

f(x) = 2.[Ln(x) / Ln(10)] - [- 2.Ln(10) / Ln(x)]

f(x) = 2.[Ln(x) / Ln(10)] + [2.Ln(10) / Ln(x)]

f(x) = 2.[{Ln(x) / Ln(10)} + {Ln(10) / Ln(x)}]

f(x) = 2.[{Ln(x)}² + {Ln(10)}²] / [Ln(10).Ln(x)] ← this is a function

...and you can obtain the minimum when the derivative is zero.

f(x) = 2.[{Ln(x)}² + {Ln(10)}²] / [Ln(10).Ln(x)]

The function f looks like 2.(u/v), so the derivative looks like: 2.[(u'.v) - (v'.u)]/v² → where:

u = {Ln(x)}² + {Ln(10)}² → u' = 2 * (1/x) * Ln(x) = (2/x).Ln(x)

v = Ln(10).Ln(x) → v' = (1/x).Ln(10)

f'(x) = 2.[(u'.v) - (v'.u)]/v²

f'(x) = 2.[(2/x).Ln(x).Ln(10).Ln(x) - (1/x).Ln(10).[{Ln(x)}² + {Ln(10)}²]] / [Ln(10).Ln(x)]²

f'(x) = 2.[(2/x).{Ln(x)}².Ln(10) - (1/x).Ln(10).[{Ln(x)}² + {Ln(10)}²]] / [Ln(10).Ln(x)]²

f'(x) = (2/x).Ln(10).[2.{Ln(x)}²- [{Ln(x)}² + {Ln(10)}²]] / [Ln(10).Ln(x)]²

f'(x) = (2/x).Ln(10).[2.{Ln(x)}²- {Ln(x)}² - {Ln(10)}²] / [Ln(10).Ln(x)]²

f'(x) = (2/x).Ln(10).[{Ln(x)}² - {Ln(10)}²] / [Ln(10).Ln(x)]²

The you must solve for x, the equation: f'(x) = 0

(2/x).Ln(10).[{Ln(x)}² - {Ln(10)}²] / [Ln(10).Ln(x)]² = 0 → recall: x > 1 → the denominator can be zero

(2/x).Ln(10).[{Ln(x)}² - {Ln(10)}²] = 0 → recall: x > 1 → so(x/2) ≠ 0

Ln(10).[{Ln(x)}² - {Ln(10)}²] = 0

{Ln(x)}² - {Ln(10)}² = 0

{Ln(x)}² = {Ln(10)}²

x = 10

Recall the function:

f(x) = 2.[{Ln(x)}² + {Ln(10)}²] / [Ln(10).Ln(x)] → when: x = 10

f(10) = 2.[Ln(10)}² + {Ln(10)}²] / [Ln(10).Ln(10)]

f(10) = 2.[2.{Ln(10)}²] / [Ln(10)]²

f(10) = 4 ← this is the least value