A Level Mathematics: Logarithmic and exponential functions, questions with answers

Rewrite each term as a single logarithm: 2ln p = ln p², ½ln(q+1) = ln√(q+1), and 3 = ln e³. So ln p² + ln(p−1) − ln√(q+1) = ln e³, giving ln[p²(p−1)/√(q+1)] = ln e³. So p²(p−1)/√(q+1) = e³, hence √(q+1) = p²(p−1)/e³. Squaring: q + 1 = p⁴(p−1)²/e⁶, so q = p⁴(p−1)²/e⁶ − 1.

Combine all terms into a single logarithm equation, remove the logs, then square to isolate q.

Let y = e²ˣ: y² − 8y − 48 = 0 → (y − 12)(y + 4) = 0 → y = 12 or y = −4.
Since e²ˣ > 0, reject y = −4. So e²ˣ = 12 → 2x = ln12 → x = ½ln12 = 1.24 (3 s.f.), and no other solution.

Substitute y = e²ˣ to form a quadratic in y, then reject any negative root since e²ˣ must be positive.

(a) 2 = log₄16, and 2log₄(3x−1) = log₄(3x−1)². So log₄(2x+1) = log₄[(3x−1)²/16], giving 2x+1 = (3x−1)²/16.
16(2x+1) = (3x−1)² → 32x + 16 = 9x² − 6x + 1 → 9x² − 38x − 15 = 0.
(b) x = [38 ± √(1444+540)]/18 = [38 ± 8√31]/18 = (19 ± 4√31)/9.
x = 4.59 or x = −0.363. Since we need 3x−1 > 0 (x > 1/3), reject x = −0.363. So x = 4.59 only.

Use the power law for logarithms and the fact 2 = log₄16 to clear the logs, then solve the resulting quadratic and reject any root outside the domain 3x−1>0.

ln(x−5) + ln x = 7 ⇒ ln[x(x−5)] = 7 ⇒ x(x−5) = e⁷ ⇒ x² − 5x − e⁷ = 0.
x = [5 + √(25 + 4e⁷)]/2 (taking the root with x > 5, since ln(x−5) requires x > 5)
= [5 + √4411.53]/2 ≈ (5 + 66.42)/2 ≈ 35.71.
So x = 35.71 (the only valid root).

Combine the logarithms into a single log, remove the log to form a quadratic in x, then solve keeping the root with x > 5.

Multiply through by e²ˣ: 3e⁴ˣ − 5e²ˣ − 4 = 0.
Let u = e²ˣ: 3u² − 5u − 4 = 0 → u = [5 ± √(25+48)]/6 = [5 ± √73]/6.
Since u > 0, u = (5+√73)/6 = 2.2573.
2x = ln(2.2573) = 0.8144 → x = 0.4072
x = 0.407

Multiply by e^(2x) to form a quadratic in e^(2x), solve for the positive root, then take logs.

2(3^(2x-1)) = 4^(x+1)
Take natural logs of both sides:
ln2 + (2x−1)ln3 = (x+1)ln4
ln2 + 2x ln3 − ln3 = x ln4 + ln4
x(2ln3 − ln4) = ln4 + ln3 − ln2
x = (ln4 + ln3 − ln2)/(2ln3 − ln4)
x = (1.386294 + 1.098612 − 0.693147)/(2.197224 − 1.386294)
x = 1.791759/0.810930
x = 2.2094...
x = 2.21 (2 d.p.)

Take logarithms of both sides to turn the exponential equation into a linear equation in x, then solve for x.

Taking logarithms of both sides: (3x - 1)ln2 = ln5 - x ln3. Expanding: 3x ln2 - ln2 = ln5 - x ln3 → x(3ln2 + ln3) = ln5 + ln2 → x(ln8 + ln3) = ln10 → x ln24 = ln10. So x = ln10/ln24.

Take natural logs of both sides, use the power and product laws, and collect the x terms to solve for x.

(2eˣ + e⁻ˣ)/(2 + eˣ) = 3
2eˣ + e⁻ˣ = 3(2 + eˣ) = 6 + 3eˣ
e⁻ˣ − eˣ − 6 = 0
Multiply through by eˣ:
1 − e²ˣ − 6eˣ = 0 → e²ˣ + 6eˣ − 1 = 0
Let u = eˣ:
u² + 6u − 1 = 0
u = (−6 ± √(36 + 4))/2 = (−6 ± √40)/2 = −3 ± √10
√10 ≈ 3.16228, so u = −3 + √10 ≈ 0.16228 or u = −3 − √10 ≈ −6.16228
Since u = eˣ > 0 for all real x, the root u = −3 − √10 is impossible and must be rejected. This shows the equation has only one real root.
eˣ = √10 − 3
x = ln(√10 − 3) = −1.818 (3 d.p.)

Multiply out and substitute u = eˣ to get a quadratic in u, then reject the negative root because eˣ can never be negative.