If x, y, z are three consecutive positive integers, then log (1 + xz)

1.	If x, y, z are three consecutive positive integers, then log (1 + xz) is1. log y2. \(\log \dfrac{y}{2}\)3. log (2y)4. 2 log (y)
Answer» Correct Answer - Option 4 : 2 log (y) Concept: Logarithm Rule log mⁿ= n log m Calculations: Let x, y, z are three consecutive positive integers. ⇒ y = x + 1 and z = y + 1 ⇒ z = x + 2 Consider, log (1 + xz) = log [1 + x(x+2)] = log [1 + x²+ 2x] = log (1 + x)² = 2 log (1 + x) = 2 log y Hence, If x, y, z are three consecutive positive integers, then log (1 + xz) is 2 log y

If x, y, z are three consecutive positive integers, then log (1 + xz) is1. log y2. \(\log \dfrac{y}{2}\)3. log (2y)4. 2 log (y)

Answer» Correct Answer - Option 4 : 2 log (y)

Concept:

Logarithm Rule

log mⁿ= n log m

Calculations:

Let x, y, z are three consecutive positive integers.

⇒ y = x + 1 and z = y + 1

⇒ z = x + 2

Consider, log (1 + xz)

= log [1 + x(x+2)]

= log [1 + x²+ 2x]

= log (1 + x)²

= 2 log (1 + x)

= 2 log y

Hence, If x, y, z are three consecutive positive integers, then log (1 + xz) is 2 log y

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If x, y, z are three consecutive positive integers, then log (1 + xz) is1. log y2. \(\log \dfrac{y}{2}\)3. log (2y)4. 2 log (y)

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