TS ICET 24 May 2019 Shift 1 Paper

Question : 77 of 200

Marks: +1, -0

3^a

5^b

75^c

⇒ c(2a + b) =

Solution:

3^{a}=5^{b}=75^{c}=k

⇒ 3=K^{\;{1}/{a}}, 5=K^{\;{1}/{3}}, 75=K^{\;{1}/{c}}

75=5^{2} × 3

K^{\;{1}/{c}}=(K^{\;{1}/{b}})^{2} ⋅ K^{\;{1}/{a}}=k^{\;{2}/{b}+\;{1}/{a}}

\;{1}/{c}=\;{2}/{b}+\;{1}/{a}

⇒ \;{2 a+b}/{a b}=\;{1}/{c}

⇒ c(2 a+b)=a b

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