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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