If log₁₀ 2 = 0.3010, then log₂ 10 is equal to:

a^x = b^y

⟹ log a^x = log b^y

⟹ x log a = y log b

⟹	log a	=	y	.
	log b		x

log	a	+ log	b	= log (a + b)
	b		a

⟹ log (a + b) = log	❨	a	x	b	❩	= log 1.
		b		a

So, a + b = 1.

Speed downstream = (13 + 4) km/hr = 17 km/hr.

Time taken to travel 68 km downstream =	❨	68	❩hrs = 4 hrs.
		17

Let the man's rate upstream be x kmph and that downstream be y kmph.

Then, distance covered upstream in 8 hrs 48 min = Distance covered downstream in 4 hrs.

⟹	❨	x x 8	4	❩	= (y x 4)
			5

⟹	44	x =4y
	5

⟹ y =	11	x.
	5

∴ Required ratio =	❨	y + x	❩	:	❨	y - x	❩
		2				2

=	❨	16x	x	1	❩	:	❨	6x	x	1	❩
		5		2				5		2

=	8	:	3
	5		5

   = 8 : 3.

Let man's rate upstream be x kmph.

Then, his rate downstream = 2x kmph.

∴ (Speed in still water) : (Speed of stream) =	❨	2x + x	❩	:	❨	2x - x	❩
		2				2

=	3x	:	x
	2		2

   = 3 : 1.

log10 80	= log10 (8 x 10)
	= log10 8 + log10 10
	= log10 (23 ) + 1
	= 3 log10 2 + 1
	= (3 x 0.3010) + 1
	= 1.9030.

Let log₂ 16 = n.

Then, 2ⁿ = 16 = 2⁴     ⟹     n = 4.

∴ log₂ 16 = 4.

log₁₀ 5 + log₁₀ (5x + 1) = log₁₀ (x + 5) + 1

⟹ log₁₀ 5 + log₁₀ (5x + 1) = log₁₀ (x + 5) + log₁₀ 10

⟹ log₁₀ [5 (5x + 1)] = log₁₀ [10(x + 5)]

⟹ 5(5x + 1) = 10(x + 5)

⟹ 5x + 1 = 2x + 10

⟹ 3x = 9

⟹ x = 3.

(a) Since log_a a = 1, so log₁₀ 10 = 1.

(b) log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3

     ∴ log (2 + 3) ≠ log (2 x 3)

(c) Since log_a 1 = 0, so log₁₀ 1 = 0.

(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.

So, (b) is incorrect.

log (264)	= 64 x log 2
	= (64 x 0.30103)
	= 19.26592

Its characteristic is 19.

Hence, then number of digits in 2⁶⁴ is 20.