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secondary 3 | A Maths
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Logarithms
Date Posted: 3 years ago
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5e²ˣ + 6e² = 17eˣ⁺¹
5(eˣ)² + 6e² = 17e(eˣ)
Use the substitution u = eˣ
5u² + 6e² = 17eu
5u² - 17eu + 6e² = 0
Factorise (you can use cross method or other methods. Remember that e is a constant and not a variable, but the procedure is the same)
(5u - 2e)(u - 3e) = 0
5u - 2e = 0 or u - 3e = 0
5u = 2e or u = 3e
u = 2e/5
So, substitute eˣ back into the equations,
eˣ = 2e/5 or eˣ = 3e
eˣ/e = 2/5 or eˣ/e = 3
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
5(eˣ)² + 6e² = 17e(eˣ)
Use the substitution u = eˣ
5u² + 6e² = 17eu
5u² - 17eu + 6e² = 0
Factorise (you can use cross method or other methods. Remember that e is a constant and not a variable, but the procedure is the same)
(5u - 2e)(u - 3e) = 0
5u - 2e = 0 or u - 3e = 0
5u = 2e or u = 3e
u = 2e/5
So, substitute eˣ back into the equations,
eˣ = 2e/5 or eˣ = 3e
eˣ/e = 2/5 or eˣ/e = 3
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
Thank you
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
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