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Question Number 120120 by benjo_mathlover last updated on 29/Oct/20
 { ((log _a (x) = 8)),((log _b (x) = 3 )),((log _c (x) = 6)) :} ⇒ log _(abc)  (x)=?
$$\begin{cases}{\mathrm{log}\:_{{a}} \left({x}\right)\:=\:\mathrm{8}}\\{\mathrm{log}\:_{{b}} \left({x}\right)\:=\:\mathrm{3}\:}\\{\mathrm{log}\:_{{c}} \left({x}\right)\:=\:\mathrm{6}}\end{cases}\:\Rightarrow\:\mathrm{log}\:_{{abc}} \:\left({x}\right)=? \\ $$
Answered by bemath last updated on 29/Oct/20
 { ((log _a (x)=8⇒log _x (a)=(1/8))),((log _b (x)=3⇒ log _x (b)=(1/3))),((log _c (x)=6⇒log _x (c)=(1/6))) :}  Then log _(abc) (x) = (1/(log _x (abc))) = (1/((1/( 8))+(1/3)+(1/6)))                        = ((24)/(3+8+4)) = ((24)/(15)) = (8/5)
$$\begin{cases}{\mathrm{log}\:_{{a}} \left({x}\right)=\mathrm{8}\Rightarrow\mathrm{log}\:_{{x}} \left({a}\right)=\frac{\mathrm{1}}{\mathrm{8}}}\\{\mathrm{log}\:_{{b}} \left({x}\right)=\mathrm{3}\Rightarrow\:\mathrm{log}\:_{{x}} \left({b}\right)=\frac{\mathrm{1}}{\mathrm{3}}}\\{\mathrm{log}\:_{{c}} \left({x}\right)=\mathrm{6}\Rightarrow\mathrm{log}\:_{{x}} \left({c}\right)=\frac{\mathrm{1}}{\mathrm{6}}}\end{cases} \\ $$$${Then}\:\mathrm{log}\:_{{abc}} \left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{log}\:_{{x}} \left({abc}\right)}\:=\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{24}}{\mathrm{3}+\mathrm{8}+\mathrm{4}}\:=\:\frac{\mathrm{24}}{\mathrm{15}}\:=\:\frac{\mathrm{8}}{\mathrm{5}} \\ $$
Answered by $@y@m last updated on 29/Oct/20
x=a^8 =b^3 =c^6   a=x^(1/8) ,  b=x^(1/3) , c=x^(1/6)   abc=x^((1/8)+(1/3)+(1/6)) =x^((15)/(24))   log_x  (abc)=((15)/(24))  log _(abc)  (x)=((24)/(15))=(8/5)
$${x}={a}^{\mathrm{8}} ={b}^{\mathrm{3}} ={c}^{\mathrm{6}} \\ $$$${a}={x}^{\frac{\mathrm{1}}{\mathrm{8}}} ,\:\:{b}={x}^{\frac{\mathrm{1}}{\mathrm{3}}} ,\:{c}={x}^{\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$${abc}={x}^{\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}} ={x}^{\frac{\mathrm{15}}{\mathrm{24}}} \\ $$$$\mathrm{log}_{{x}} \:\left({abc}\right)=\frac{\mathrm{15}}{\mathrm{24}} \\ $$$$\mathrm{log}\:_{{abc}} \:\left({x}\right)=\frac{\mathrm{24}}{\mathrm{15}}=\frac{\mathrm{8}}{\mathrm{5}} \\ $$

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