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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...

$${most}\:{welcome}... \\ $$

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