Question

How do you solve for x in `1/3lnx+ln2-ln3=3`?

Answer 1

Hello !

Write `\ln(x) = 3\times (3 + \ln(3) - ln(2))` and apply exp function :

`x = e^{3\times (3 + \ln(3) - ln(2))} = e^9\times e^{3 ln(3)}\times e^{-3 \ln(2)}`. You can simplify :

`x = e^9\times 27 \times \frac{1}{8} = \frac{27}{8}e^9`,

Remark. I used the rules
1) `e^{-x} = \frac{1}{e^x}`.
2) `n ln(a) = ln(a^n)`.
3) `e^{ln (a)} = a` if `a >0`.

Answer 2

The answer is: `x=27/8e^9`.

First of all we have to add a condition otherwise our equation loses meaning: `x>0`.

Than:

`1/3lnx+ln2-ln3=3rArrlnx=3(ln3-ln2+3)rArr`

`lnx=3(ln3-ln2+lne^3)rArrlnx=3ln(3e^3/2)rArr`

`lnx=ln(27/8e^9)rArrx=27/8e^9`

(That is positive, so it is acceptable).