જો {{log x} over {b - c}} = {{log y} over {c | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 =

Vedclass · Accepted Answer

ઉપરોક્ત બધાજ

Vedclass · Answer

(d) ${{\log x} \over {b - c}} = {{\log y} \over {c - a}} = {{\log z} \over {a - b}} = k\,({m{say}})$

$ \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)}}$

$	herefore xyz = {e^{k(b - c) + k(c - a) + k(a - b)}} = {e^0} = 1$

${x^a}{y^b}{z^c} = {e^{k(b - c)a + k(c - a)b + k(a - b)c}} = {e^0} = 1 = xyz$

${x^{b + c}}{y^{c + a}}{z^{a + b}} = {e^{k({b^2} - {c^2}) + k({c^2} - {a^2}) + k({a^2} - {b^2})}} = {e^0} = 1$.

જો ${{\log x} \over {b - c}} = {{\log y} \over {c - a}} = {{\log z} \over {a - b}} $ તો આપલે પૈકી . . . સત્ય છે.

Similar Questions

${\log _2}.{\log _3}....{\log _{100}}{100^{{{99}^{{{98}^{{.^{{.^{{{.2}^1}}}}}}}}}}}= . . . $.

જો $x = {\log _5}(1000)$ અને $y = {\log _7}(2058)$ તો

જો ${a^x} = b,{b^y} = c,{c^z} = a,$ તો $xyz = . . . .$

${\log _2}7$ એ . . . . થાય.

જો $A = {\log _2}{\log _2}{\log _4}256 + 2{\log _{\sqrt 2 \,}}\,2$ તો $A = . . . .$