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solve for x

3ln(-3x+5) = -9

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

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)*

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