A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream.

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.

Speed in still water =	1	(11 + 5) kmph = 8 kmph.
	2

Speed upstream = 7.5 kmph.

Speed downstream = 10.5 kmph.

∴ Total time taken =	❨	105	+	105	❩hours = 24 hours.
		7.5		10.5

Suppose he move 4 km downstream in x hours. Then,

Speed downstream =	❨	4	❩	km/hr.
		x

Speed upstream =	❨	3	❩	km/hr.
		x

∴	48	+	48	= 14 or x =	1	.
	(4/x)		(3/x)		2

So, Speed downstream = 8 km/hr, Speed upstream = 6 km/hr.

Rate of the stream =	1	(8 - 6) km/hr = 1 km/hr.
	2

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

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

So, a + b = 1.

a^x = b^y

⟹ log a^x = log b^y

⟹ x log a = y log b

⟹	log a	=	y	.
	log b		x

log2 10 =	1	=	1	=	10000	=	1000	.
	log10 2		0.3010		3010		301

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.