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Question Number 156676 by mathlove last updated on 14/Oct/21

Commented by MathSh last updated on 14/Oct/21

(2/a) + (1/b) + (1/c) = 3 ⇒ ((21)/3) = 7

$$\frac{\mathrm{2}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:+\:\frac{\mathrm{1}}{\mathrm{c}}\:=\:\mathrm{3}\:\Rightarrow\:\frac{\mathrm{21}}{\mathrm{3}}\:=\:\mathrm{7} \\ $$

Answered by som(math1967) last updated on 14/Oct/21

60=27^a ⇒60^(1/a) =27  60^(1/b) =64    ,60^(1/c) =125  60^((1/a)+(1/b)+(1/c)) =3^3 ×4^3 ×5^3   (60)^((ab+bc+ca)/(abc)) =(60)^3    ((ab+bc+ca)/(abc))=3  ((abc)/(ab+bc+ca))=(1/3)  ∴((21abc)/(ab+bc+ca))=21×(1/3)=7 ans

$$\mathrm{60}=\mathrm{27}^{{a}} \Rightarrow\mathrm{60}^{\frac{\mathrm{1}}{{a}}} =\mathrm{27} \\ $$$$\mathrm{60}^{\frac{\mathrm{1}}{{b}}} =\mathrm{64}\:\:\:\:,\mathrm{60}^{\frac{\mathrm{1}}{{c}}} =\mathrm{125} \\ $$$$\mathrm{60}^{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}} =\mathrm{3}^{\mathrm{3}} ×\mathrm{4}^{\mathrm{3}} ×\mathrm{5}^{\mathrm{3}} \\ $$$$\left(\mathrm{60}\right)^{\frac{{ab}+{bc}+{ca}}{{abc}}} =\left(\mathrm{60}\right)^{\mathrm{3}} \\ $$$$\:\frac{{ab}+{bc}+{ca}}{{abc}}=\mathrm{3} \\ $$$$\frac{{abc}}{{ab}+{bc}+{ca}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\therefore\frac{\mathrm{21}\boldsymbol{{abc}}}{\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ca}}}=\mathrm{21}×\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{7}\:\boldsymbol{{ans}} \\ $$

Answered by Rasheed.Sindhi last updated on 14/Oct/21

          An Alternate Approach  (3^a )^3 =(4^b )^3 =(5^c )^3 =60  (3^a )^3 (4^b )^3 (5^c )^3 =(60)^3   3^a 4^b 5^c =60=3^1 .4^1 .5^1   a=b=c=1  ((21abc)/(ab+bc+ca))=((21.1.1.1)/(1.1+1.1+1.1))=((21)/3)=7

$$\:\:\:\:\:\:\:\:\:\:\mathbb{A}\mathrm{n}\:\mathbb{A}\mathrm{lternate}\:\mathbb{A}\mathrm{pproach} \\ $$$$\left(\mathrm{3}^{{a}} \right)^{\mathrm{3}} =\left(\mathrm{4}^{{b}} \right)^{\mathrm{3}} =\left(\mathrm{5}^{{c}} \right)^{\mathrm{3}} =\mathrm{60} \\ $$$$\left(\mathrm{3}^{{a}} \right)^{\mathrm{3}} \left(\mathrm{4}^{{b}} \right)^{\mathrm{3}} \left(\mathrm{5}^{{c}} \right)^{\mathrm{3}} =\left(\mathrm{60}\right)^{\mathrm{3}} \\ $$$$\mathrm{3}^{{a}} \mathrm{4}^{{b}} \mathrm{5}^{{c}} =\mathrm{60}=\mathrm{3}^{\mathrm{1}} .\mathrm{4}^{\mathrm{1}} .\mathrm{5}^{\mathrm{1}} \\ $$$${a}={b}={c}=\mathrm{1} \\ $$$$\frac{\mathrm{21}{abc}}{{ab}+{bc}+{ca}}=\frac{\mathrm{21}.\mathrm{1}.\mathrm{1}.\mathrm{1}}{\mathrm{1}.\mathrm{1}+\mathrm{1}.\mathrm{1}+\mathrm{1}.\mathrm{1}}=\frac{\mathrm{21}}{\mathrm{3}}=\mathrm{7} \\ $$$$ \\ $$

Commented by mr W last updated on 14/Oct/21

but a=b=c=1 doesn′t fulfill the  given condition.

$${but}\:{a}={b}={c}=\mathrm{1}\:{doesn}'{t}\:{fulfill}\:{the} \\ $$$${given}\:{condition}. \\ $$

Commented by Rasheed.Sindhi last updated on 14/Oct/21

Yes sir the approach is logically  wrong but the answer is by chance  right! Thanks sir!

$${Yes}\:{sir}\:{the}\:{approach}\:{is}\:\boldsymbol{{logically}} \\ $$$$\boldsymbol{{wrong}}\:{but}\:{the}\:{answer}\:{is}\:\boldsymbol{{by}}\:\boldsymbol{{chance}} \\ $$$$\boldsymbol{{right}}!\:\mathcal{T}{hanks}\:{sir}! \\ $$

Answered by john_santu last updated on 14/Oct/21

 ((21)/((1/c)+(1/a)+(1/b))) = k    { ((27^a =60⇒a=log _(27) (60))),((64^b =60⇒b=log _(64) (60))),((125^c =60⇒c=log _(125) (60))) :}   therefore k=((21)/(log _(60) (125)+log _(60) (27)+log _(60) (64)))   k=((21)/(log _(60) (125×27×64))) =((27)/(log _(60) (60^3 )))   k=((21)/(3log _(60) (60)))=((21)/3)=7.

$$\:\frac{\mathrm{21}}{\frac{\mathrm{1}}{{c}}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}}\:=\:{k} \\ $$$$\:\begin{cases}{\mathrm{27}^{{a}} =\mathrm{60}\Rightarrow{a}=\mathrm{log}\:_{\mathrm{27}} \left(\mathrm{60}\right)}\\{\mathrm{64}^{{b}} =\mathrm{60}\Rightarrow{b}=\mathrm{log}\:_{\mathrm{64}} \left(\mathrm{60}\right)}\\{\mathrm{125}^{{c}} =\mathrm{60}\Rightarrow{c}=\mathrm{log}\:_{\mathrm{125}} \left(\mathrm{60}\right)}\end{cases} \\ $$$$\:{therefore}\:{k}=\frac{\mathrm{21}}{\mathrm{log}\:_{\mathrm{60}} \left(\mathrm{125}\right)+\mathrm{log}\:_{\mathrm{60}} \left(\mathrm{27}\right)+\mathrm{log}\:_{\mathrm{60}} \left(\mathrm{64}\right)} \\ $$$$\:{k}=\frac{\mathrm{21}}{\mathrm{log}\:_{\mathrm{60}} \left(\mathrm{125}×\mathrm{27}×\mathrm{64}\right)}\:=\frac{\mathrm{27}}{\mathrm{log}\:_{\mathrm{60}} \left(\mathrm{60}^{\mathrm{3}} \right)} \\ $$$$\:{k}=\frac{\mathrm{21}}{\mathrm{3log}\:_{\mathrm{60}} \left(\mathrm{60}\right)}=\frac{\mathrm{21}}{\mathrm{3}}=\mathrm{7}.\: \\ $$

Commented by cortano last updated on 14/Oct/21

waw

$$\mathrm{waw} \\ $$$$ \\ $$

Commented by peter frank last updated on 14/Oct/21

great

$$\mathrm{great} \\ $$

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