Question

42.

Answer 1

Answer: (d) 0

we have:

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

Step-by-step explanation:

Answer 2

Answer:

abc = 0

Step-by-step explanation:

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