# pre cal

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(1/3)LN27 + 2LNx = LN(2-x)

i think i just need help getting started.

I assume that your LN notation is log to base e. I will call it ln.
(1/3)ln 27 = ln (27)^(1/3) = ln 3,
(since 27^(1/3) is the cube root of 27).
Therefore
ln 3 + 2 ln x = ln (3 x^2) = ln (2-x)
3x^2 = 2-x
solve the quadratic equation. It can be factored.

The answer would be the same, regardless of the log base.

thank you!!! yes, that is what i meant, but when i typed in lower case "l" it looked like a one and i didn't want anyone to be mistaken. thanks for your help!!

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