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

Join Wrath of Math to get exclusive videos, music, and more:

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

◉Textbooks I Like◉

★DONATE★

Thanks to Loke Tan, Raül Beienheimer, Matt Venia, Micheline, Doug Walker, Odd Hultberg, Marc, Shlome Ashkenazi, Barbora Sharrock, Mohamad Nossier, Rolf Waefler, Shadow Master, and James Mead for their generous support on Patreon!

Outro music is mine. You cannot find it anywhere, for now.

Follow Wrath of Math on...

[supreme--qwerty]

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]

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