J's answer to Beth's Secondary 3 A Maths Singapore question.
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
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
Date Posted:
3 years ago
Thank you
360 Tutoring Program