Question

42.

*=3"y=9,- 27 and "y"? - 1, then abe is a
(a) 1
) (b) 27
(e)
( (d) 0
43. In given fig., the diagonals of a rectangle ABCD​

Answer 1

abc = 0

Explanation:

`x = {3}^(a) \\ y = {9}^(b) = {3}^(2b) \\ z = {27}^(c) = {3}^(3c) \\ \\ \because \: {x}^(bc) {y}^(ca) {z}^(ab) = 1 \\ \\ \therefore \: ( {3}^(a) )^(bc) ( {3}^(2b) )^(ca) ( {3}^(3c) )^(ab) = 1 \\ \\ {3}^(abc) .{3}^(2abc) .{3}^(3abc) = 1 \\ \\ {3}^(abc + 2abc + 3abc) = 1 \\ \\ {3}^(6abc) = {3}^(0) \: \\ ( \because \: {3}^(0) = 1) \\ \\ \because \: bases \: are \: equal \\ \therefore \: exponents \: will \: also \: be \: equal \\ \\ \implies \: 6abc = 0 \\ \\ \implies \: \red{ \bold{abc = 0 }}`

Answer 2

Answer: (d) 0

we have:

`y=9^(b)=(3^(2))^(b)=3^(2b)\\\\z=27^(c)=(3^(3))^(c)=3^(3c)\\\\\\x^(bc).y^(ca).z^(ab)=1\\\\<=>(3^(a))^(bc).(3^(2b))^(ca).(3^(3c))^(ab)=1\\\\\\<=>3^(abc).3^(2abc).3^(3abc)=1\\\\\\<=>3^(6abc)=3^(0)\\\\\\=>6abc=0\\\\\\<=>abc=0`

Explanation: