Menu Close

If-xyz-0-and-x-y-z-0-a-10-z-b-10-y-c-10-x-then-a-1-y-1-z-b-1-z-1-x-c-1-x-1-y-a-0-001-b-0-01-c-0-1-d-1-e-10-

Tinku Tara
Pin




Question Number 65781 by gunawan last updated on 03/Aug/19
If xyz ≠ 0 and x+y+z=0 a=10^z b=10^y c=10^x then a^(((1/y)+(1/z))) . b^(((1/z)+(1/x))) .c^(((1/x)+(1/y))) =... a. 0.001 b. 0.01 c. 0.1 d. 1 e. 10
$$\mathrm{If}\:{xyz}\:\neq\:\mathrm{0}\:\mathrm{and}\:{x}+{y}+{z}=\mathrm{0} \\ $$$${a}=\mathrm{10}^{{z}} \\ $$$${b}=\mathrm{10}^{{y}} \\ $$$${c}=\mathrm{10}^{{x}} \\ $$$$\mathrm{then} \\ $$$${a}^{\left(\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)} .\:{b}^{\left(\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{x}}\right)} .{c}^{\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\right)} =… \\ $$$${a}.\:\mathrm{0}.\mathrm{001} \\ $$$${b}.\:\mathrm{0}.\mathrm{01} \\ $$$${c}.\:\mathrm{0}.\mathrm{1} \\ $$$${d}.\:\mathrm{1} \\ $$$${e}.\:\mathrm{10} \\ $$$$ \\ $$
Answered by Tanmay chaudhury last updated on 03/Aug/19
p=a^(((1/y)+(1/z))) .b^(((1/z)+(1/x))) .c^(((1/x)+(1/(y )))) log_(10) p=((1/y)+(1/z))log_(10) a+((1/z)+(1/x))log_(10) b+((1/x)+(1/y))log_(10) c log_(10) p=((1/y)+(1/z))log_(10) 10^x +((1/z)+(1/x))log_(10) 10^y +((1/x)+(1/y))log_(10) 10^z =((x/y)+(x/z))+((y/z)+(y/x))+((z/x)+(z/y)) =((x+z)/y)+((x+y)/z)+((y+z)/x) =((x+z)/y)+1+((x+y)/z)+1+((y+z)/x)+1−3 =(x+y+z)((1/x)+(1/y)+(1/z))−3 =−3 so log_(10) p=−3 p=10^(−3) =(1/(1000))=0.001
$${p}={a}^{\left(\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)} .{b}^{\left(\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{x}}\right)} .{c}^{\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}\:}\right)} \\ $$$${log}_{\mathrm{10}} {p}=\left(\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right){log}_{\mathrm{10}} {a}+\left(\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{x}}\right){log}_{\mathrm{10}} {b}+\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\right){log}_{\mathrm{10}} {c} \\ $$$${log}_{\mathrm{10}} {p}=\left(\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right){log}_{\mathrm{10}} \mathrm{10}^{{x}} +\left(\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{x}}\right){log}_{\mathrm{10}} \mathrm{10}^{{y}} +\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\right){log}_{\mathrm{10}} \mathrm{10}^{{z}} \\ $$$$=\left(\frac{{x}}{{y}}+\frac{{x}}{{z}}\right)+\left(\frac{{y}}{{z}}+\frac{{y}}{{x}}\right)+\left(\frac{{z}}{{x}}+\frac{{z}}{{y}}\right) \\ $$$$=\frac{{x}+{z}}{{y}}+\frac{{x}+{y}}{{z}}+\frac{{y}+{z}}{{x}} \\ $$$$=\frac{{x}+{z}}{{y}}+\mathrm{1}+\frac{{x}+{y}}{{z}}+\mathrm{1}+\frac{{y}+{z}}{{x}}+\mathrm{1}−\mathrm{3} \\ $$$$=\left({x}+{y}+{z}\right)\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)−\mathrm{3} \\ $$$$=−\mathrm{3} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{log}}_{\mathrm{10}} \boldsymbol{{p}}=−\mathrm{3} \\ $$$$\boldsymbol{{p}}=\mathrm{10}^{−\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{1000}}=\mathrm{0}.\mathrm{001} \\ $$$$ \\ $$
Commented by gunawan last updated on 03/Aug/19
wow thank you Sir
$$\mathrm{wow} \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$
Commented by Tanmay chaudhury last updated on 04/Aug/19
most welcome...
$${most}\:{welcome}… \\ $$

Pin