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secondary 3 | A Maths

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log₅(x²-5x+20) = 1/log₂5 + 1/log₇5

=> log₅(x²-5x+20) = log2/log5 + log7/log5

=> log₅(x²-5x+20) = log₅2 + log₅7

=> log₅(x²-5x+20) = log₅14

=> x²-5x+20 = 14

=> x²-5x+6 = 0

=> (x-2)(x-3) = 0

=> x = 2 or 3

(a)(ii)

lg y² = (lg y)²

=> 2lg y = (lg y)²

=> (lg y)² - 2lg y = 0

=> lg y(lg y-2) = 0

=> lg y = 0 or lg y - 2 = 0

=> y = 1 or lg y = 2

=> y = 1 or y = 100

(b)

8e⁴ˣ - 15e²ˣ - 2 = 0

=> 8(e²ˣ)² - 15e²ˣ - 2 = 0

=> (8e²ˣ + 1)(e²ˣ - 2) = 0

=> e²ˣ = -1/8 (n.a.) or e²ˣ = 2

=> e²ˣ = 2

=> 2x = ln 2

=> x = (ln 2)/2 = 0.347

...

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I hope my explanations and workings help! Let me know if you still need more explanation for my workings and I will explain them again!