J's answer to Beth's Secondary 3 A Maths Singapore question.
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Alternatively,
5e²ˣ + 6e² = 17eˣ⁺¹
Divide both sides by e² so we can get 6 for the constant.
5e²ˣ / e² + 6 = 17eˣ⁺¹ / e²
5e²ˣ⁻² + 6 = 17eˣ⁺¹⁻²
5e²ˣ⁻² + 6 = 17eˣ⁻¹
5(eˣ⁻¹)² - 17eˣ⁻¹ + 6 = 0
Then use the substitution u = eˣ⁻¹
5u² - 17u + 6 = 0
Factorise,
(5u - 2)(u - 3) = 0
5u - 2 = 0 or u - 3 = 0
5u = 2 or u = 3
u = 2/5
So, substitute eˣ⁻¹ back into the equations,
eˣ⁻¹ = 2/5 or eˣ⁻¹ = 3
x - 1 = ln (2/5) or x - 1 = ln 3
x = ln (2/5) + 1 or x = ln 3 + 1
x = ln 2 - ln 5 + 1
5e²ˣ + 6e² = 17eˣ⁺¹
Divide both sides by e² so we can get 6 for the constant.
5e²ˣ / e² + 6 = 17eˣ⁺¹ / e²
5e²ˣ⁻² + 6 = 17eˣ⁺¹⁻²
5e²ˣ⁻² + 6 = 17eˣ⁻¹
5(eˣ⁻¹)² - 17eˣ⁻¹ + 6 = 0
Then use the substitution u = eˣ⁻¹
5u² - 17u + 6 = 0
Factorise,
(5u - 2)(u - 3) = 0
5u - 2 = 0 or u - 3 = 0
5u = 2 or u = 3
u = 2/5
So, substitute eˣ⁻¹ back into the equations,
eˣ⁻¹ = 2/5 or eˣ⁻¹ = 3
x - 1 = ln (2/5) or x - 1 = ln 3
x = ln (2/5) + 1 or x = ln 3 + 1
x = ln 2 - ln 5 + 1
Date Posted:
3 years ago
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