# 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-

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