- Home
- Standard 11
- Mathematics
Basic of Logarithms
medium
If ${{\log x} \over {b - c}} = {{\log y} \over {c - a}} = {{\log z} \over {a - b}},$ then which of the following is true
A
$xyz = 1$
B
${x^a}{y^b}{z^c} = 1$
C
${x^{b + c}}{y^{c + a}}{z^{a + b}} = 1$
D
All of These
Solution
(d) ${{\log x} \over {b – c}} = {{\log y} \over {c – a}} = {{\log z} \over {a – b}} = k\,({\rm{say}})$
$ \Rightarrow $$\log x = k(b – c),\,\log y = k(c – a),\,\log z = k(a – b)$
$ \Rightarrow $$x = {e^{k(b – c)}},\,y = {e^{k(c – a)}},\,z = {e^{k(a – b)}}$
$\therefore xyz = {e^{k(b – c) + k(c – a) + k(a – b)}} = {e^0} = 1$
${x^a}{y^b}{z^c} = {e^{k(b – c)a + k(c – a)b + k(a – b)c}} = {e^0} = 1 = xyz$
${x^{b + c}}{y^{c + a}}{z^{a + b}} = {e^{k({b^2} – {c^2}) + k({c^2} – {a^2}) + k({a^2} – {b^2})}} = {e^0} = 1$.
Standard 11
Mathematics
