You are not logged in.
Pages: 1
solve for x
3ln(-3x+5) = -9
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.
Offline
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 (2005-11-29 11:03:22)
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
Offline
Pages: 1