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

Feb 19, 2015

Hello !

Write $\setminus \ln \left(x\right) = 3 \setminus \times \left(3 + \setminus \ln \left(3\right) - \ln \left(2\right)\right)$ and apply exp function :

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

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

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

Feb 19, 2015

The answer is: $x = \frac{27}{8} {e}^{9}$.

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

Than:

$\frac{1}{3} \ln x + \ln 2 - \ln 3 = 3 \Rightarrow \ln x = 3 \left(\ln 3 - \ln 2 + 3\right) \Rightarrow$

$\ln x = 3 \left(\ln 3 - \ln 2 + \ln {e}^{3}\right) \Rightarrow \ln x = 3 \ln \left(3 {e}^{3} / 2\right) \Rightarrow$

$\ln x = \ln \left(\frac{27}{8} {e}^{9}\right) \Rightarrow x = \frac{27}{8} {e}^{9}$

(That is positive, so it is acceptable).