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Question-206104




Question Number 206104 by cortano21 last updated on 07/Apr/24
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Answered by dimentri last updated on 07/Apr/24
  7^(−a) = 3^(−b) = 2^(−c) =5^(−d) = m     7=m^(−(1/a))  ,  3=m^(−(1/b))  , 2=m^(−(1/c))  , 5^(−(1/d))     210^y  = m^(−y((1/a)+(1/b)+(1/c)+(1/d))) = m    (1/y) =−((1/a)+(1/b)+(1/c)+(1/d))     (1/y)+(1/a)+(1/b)+(1/c)+(1/d)=0
$$\:\:\mathrm{7}^{−\mathrm{a}} =\:\mathrm{3}^{−\mathrm{b}} =\:\mathrm{2}^{−\mathrm{c}} =\mathrm{5}^{−\mathrm{d}} =\:\mathrm{m} \\ $$$$\:\:\:\mathrm{7}=\mathrm{m}^{−\frac{\mathrm{1}}{\mathrm{a}}} \:,\:\:\mathrm{3}=\mathrm{m}^{−\frac{\mathrm{1}}{\mathrm{b}}} \:,\:\mathrm{2}=\mathrm{m}^{−\frac{\mathrm{1}}{\mathrm{c}}} \:,\:\mathrm{5}^{−\frac{\mathrm{1}}{\mathrm{d}}} \\ $$$$\:\:\mathrm{210}^{\mathrm{y}} \:=\:\mathrm{m}^{−\mathrm{y}\left(\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}+\frac{\mathrm{1}}{\mathrm{d}}\right)} =\:\mathrm{m} \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{y}}\:=−\left(\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}+\frac{\mathrm{1}}{\mathrm{d}}\right) \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{y}}+\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}+\frac{\mathrm{1}}{\mathrm{d}}=\mathrm{0} \\ $$
Answered by MATHEMATICSAM last updated on 07/Apr/24
210^y  = 7^(−a)  = 3^(−b)  = 2^(−c)  = 5^(−d)  = k (Let)  ⇒ 210 = k^(1/y)   ⇒ 7 = k^(− (1/a))   ⇒ 3 = k^(− (1/b))   ⇒ 2 = k^(− (1/c))   ⇒ 5 = k^(− (1/d))     210 = 7 × 3 × 2 × 5  ⇒ k^(1/y)  = k^(− (1/a))  × k^(− (1/b))  × k^(− (1/c))  × k^(− (1/d))   ⇒ k^(1/y)  = k^(− (1/a) − (1/b) − (1/c) − (1/d))   ⇒ (1/y) = − (1/a) − (1/b) − (1/c) − (1/d)   ⇒ (1/y) + (1/a) + (1/b) + (1/c) + (1/d) = 0 (Ans)
$$\mathrm{210}^{{y}} \:=\:\mathrm{7}^{−{a}} \:=\:\mathrm{3}^{−{b}} \:=\:\mathrm{2}^{−{c}} \:=\:\mathrm{5}^{−{d}} \:=\:{k}\:\left(\mathrm{Let}\right) \\ $$$$\Rightarrow\:\mathrm{210}\:=\:{k}^{\frac{\mathrm{1}}{{y}}} \\ $$$$\Rightarrow\:\mathrm{7}\:=\:{k}^{−\:\frac{\mathrm{1}}{{a}}} \\ $$$$\Rightarrow\:\mathrm{3}\:=\:{k}^{−\:\frac{\mathrm{1}}{{b}}} \\ $$$$\Rightarrow\:\mathrm{2}\:=\:{k}^{−\:\frac{\mathrm{1}}{{c}}} \\ $$$$\Rightarrow\:\mathrm{5}\:=\:{k}^{−\:\frac{\mathrm{1}}{{d}}} \\ $$$$ \\ $$$$\mathrm{210}\:=\:\mathrm{7}\:×\:\mathrm{3}\:×\:\mathrm{2}\:×\:\mathrm{5} \\ $$$$\Rightarrow\:{k}^{\frac{\mathrm{1}}{{y}}} \:=\:{k}^{−\:\frac{\mathrm{1}}{{a}}} \:×\:{k}^{−\:\frac{\mathrm{1}}{{b}}} \:×\:{k}^{−\:\frac{\mathrm{1}}{{c}}} \:×\:{k}^{−\:\frac{\mathrm{1}}{{d}}} \\ $$$$\Rightarrow\:{k}^{\frac{\mathrm{1}}{{y}}} \:=\:{k}^{−\:\frac{\mathrm{1}}{{a}}\:−\:\frac{\mathrm{1}}{{b}}\:−\:\frac{\mathrm{1}}{{c}}\:−\:\frac{\mathrm{1}}{{d}}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{y}}\:=\:−\:\frac{\mathrm{1}}{{a}}\:−\:\frac{\mathrm{1}}{{b}}\:−\:\frac{\mathrm{1}}{{c}}\:−\:\frac{\mathrm{1}}{{d}}\: \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:+\:\frac{\mathrm{1}}{{d}}\:=\:\mathrm{0}\:\left(\boldsymbol{\mathrm{Ans}}\right) \\ $$

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