Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
You are not logged in.
#2 2005-11-29 11:48:28
- mathsyperson
- Moderator
Offline
Re: ln solve for x
3ln(-3x+5) = -9
Using alnb = lnb^a : ln(-3x+5)³ = -9
Raise to the power of e: (-3x+5)³ = e^-9
Cube root: -3x+5 = e^-3
Solve: -3x = e^-3 - 5
x = -(e^-3 - 5)/3
Why did the vector cross the road?
It wanted to be normal.
#3 2005-11-30 10:02:50
- irspow
- Power Member
Offline
Re: ln solve for x
I came up with the same answer, but was a little puzzled at mathsyperson's approach. I am always amazed at the different approaches individuals will take. I did it like this.
3ln(-3x+5) = -9 divide by 3
ln(-3x+5) = -3 raise to e
-3x+5 = e^-3 subtract 5
-3x = e^-3 - 5 divide by -3
x = (e^-3 - 5) / -3
Last edited by irspow (2005-11-30 10:03:22)