Logarithm: Practice Problems

Sol : Option B
Here x, y and z are the sides of a right angled triangle, so z² = x² + y².

Sol : Option A
log₂ 2 + log₂ 2² + log₂ 2³ + ........ + log₂ 2ⁿ
log₂ 2 + 2log₂ 2 + 3log₂ 2 + ........ + nlog₂ 2
1+2+3+........+n
n(n+1)/2

Sol : Option B
log₅ 16, log₅ (3^x-4), log₅ (3^x+97/16) are in arithmetic progression

Sol : Option B
We have (log₂ x) ² – log₂ x⁴ - 32 = 0.
⇒ (log₂ x) ² – 4log₂ x - 32 = 0......(1)
Let log₂ x = y
(i) ⇒ y² – 4y – 32 = 0
⇒ y² – 8y + 4y – 32 = 0
⇒ y (y – 8) + 4 (y – 8) = 0
⇒ (y – 8) (y + 4) = 0
⇒ y = 8, -4
⇒ log₂ x = 8 or log₂ x = - 4
⇒ x = 2⁸ = 256 or x = 2^-4 = 1/16
Since ‘x’ is an integer so x = 256.

Sol : Option A
We have log₅y – logsub>5√y = 2 log_y 5