Question

How do you solve `ln e^x -ln e^3 = ln e^9`?

Answer 1

If you use the properties of logarithms, you will need to refer to these:

1. `lna^b = blna`
2. `lne = 1`

OR

`lne^z = z`

where `a` and `b` are constants, `e = 2.718281828cdots`, and `z` is either a constant or a variable. Therefore, we can do this:

`lne^x - lne^3 = lne^9`

`lne^x = lne^9 + lne^3`

`xcancel(lne) = 9cancel(lne) + 3cancel(lne)`

`color(blue)(x) = 9 + 3 = color(blue)(12)`