3 a = 5 b = 1 5
If a and b are numbers satisfying the equation above, find a 1 + b 1 .
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What if the question ask for a + b ?
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a + b = lo g 3 lo g 3 + lo g 5 + lo g 5 lo g 3 + lo g 5
A technical way to get it is using logarithms, but I unknown if there is an algebraic solution for it
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No matter what you use but i want the answer.
a = lo g 3 1 5 , b = lo g 5 1 5 .
a 1 + b 1 = lo g 1 5 3 + lo g 1 5 5 = lo g 1 5 1 5 = 1 .
@Abhay Kumar here is an answer to your question
a + b = lo g 3 1 5 + lo g 5 1 5 ≈ 4 . 1 4 7 6
Good solution! :)
3 a = 1 5 ⟹ ln 3 a = ln 1 5 ⟹ a = ln 3 ln 1 5 ⟹ a 1 = ln 1 5 ln 3
5 b = 1 5 ⟹ ln 5 b = ln 1 5 ⟹ b = ln 5 ln 1 5 ⟹ b 1 = ln 1 5 ln 5
a 1 + b 1 = ln 1 5 ln 3 + ln 1 5 ln 5 = 1 . 0 0
3 a = 1 5
3 a − 1 = 5
( 3 a − 1 ) b = 1 5
⟹ a = b ( a − 1 )
a + b = a b
a b a + b = 1
a 1 + b 1 = 1
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Relevant wiki: Properties of Logarithms - Basic
3 a = 1 5 ⇒ 3 = 1 5 a 1
5 b = 1 5 ⇒ 5 = 1 5 b 1
3 × 5 = 1 5 a 1 + b 1 = 1 5 ∴ a 1 + b 1 = 1 .