How to Solve Natural Log Equations | Precalculus Exercises

We go over how to solve natural log equations using the inverse of the natural log, the one to one property, and good old fashioned algebra. #precalculus
Precalculus course: • Precalculus
Precalculus Exercises playlist: • Precalculus Exercises
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0:00 Intro
0:12 lnx-ln2=0
0:45 lnx=5
1:27 ln(4x+3)=ln(2x+9)
2:32 lnx-ln(x+1)=2
5:33 ln(x+1)+ln(x+5)=ln(x-1)
7:10 Conclusion
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[@WrathofMath]

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[@supreme--qwerty6127]

For The 4th one, I solved it like this:- ln[x]-ln[x+1] = 2 ln[x/(x+1)] = 2 e^2 = x/(x+1) Taking e ≈ 2.71 (2.71)^2 ≈ x/(x+1) 7.34 ≈ x/(x+1) 7.34x + 7.34 ≈ x 6.34x + 7.34 ≈ 0 x ≈ -7.34/6.34 x ≈ -1.15, correct me if I'm wrong.

[@Nikioko]

It's actually closer to −1,16. 😉 And ln(−1,16) − ln(−0,16) = (ln(1,16) + πi) − (ln(−0,16) + πi) = ln(1,16) − ln(0,16) = 2.

[@Nikioko]

Of course, 4) and 5) have solutions. Keep in mind that ln(−x) = ln(x) + πi, and πi luckily cancels out in both cases.

[@Nikioko]

1) ln(x) − ln(2) = 0 ln(x) = ln(2) x = 2 2) ln(x) = 5 x = e⁵ ≈ 148,41 3) ln(4x + 3) = ln(2x + 9) 4x + 3 = 2x + 9 2x = 6 x = 3 4) ln(x) − ln(x + 1) = 2 x / (x + 1) = e² x = (x + 1) e² x = xe² + e² x − xe² = e² x (1 − e²) = e² x = e² / (1 − e²) ≈ −1,157 No real solution, but one complex solution: ln(−1,157) − ln(−0,157) = 2 ln(1,157) + πi − ln(0,157) + πi = 2 ln(1,157) − ln(0,157) = 2 2 = 2 5) ln(x + 1) + ln(x + 5) = ln(x − 1) (x + 1) (x + 5) = x − 1 x² + 6x + 5 = x − 1 x² + 5x + 6 = 0 x₁,₂ = −5/2 ± √(25/4 − 6) = −5/2 ± 1/2 x₁ = −2 ∨ x₂ = −3 No real solution, but two complex solutions: ln(−2 + 1) + ln(−2 + 5) = ln(−2 − 1) ln(−1) + ln(3) = ln(−3) ln(1) + πi + ln(3) = ln(3) + πi 0 + πi + ln(3) = ln(3) + πi ln(3) = ln(3) 3 = 3 ln(−3 + 1) + ln(−3 + 5) = ln(−3 − 1) ln(−2) + ln(2) = ln(−4) ln(2) + πi + ln(2) = ln(4) + πi 2 ln(2) = ln(4) ln(2²) = ln(4) 2² = 4