how do you solve for x in this equation...ln (9-6 x) - 2ln x = ln 3

Question

aum · Answer

Make use of the following log properties:
$\log(x)^n = n * \log(x)$

$\log(A) - \log(B) = \log(A/B)$

If $\log(A) = \log(B)$ then $A=B$

aum · Answer

$$
\ln (9-6 x) - 2\ln x = \ln 3 \
\ln (9-6 x) - \ln x^2 = \ln 3 \
\ln((9-6x) / x^2) = \ln(3) \
(9-6x) / x^2 = 3 \
9-6x = 3x^2  \
3x^2 + 6x - 9 = 0 \
x^2 + 2x - 3 = 0 \
$$Can you solve it now?

anonymous · Answer

Yeah i got it thanks!

aum · Answer

You are welcome.

Remember you cannot take logarithm of a negative number.  If you get an x value that will end up taking log of a negative number discard that solution as extraneous solution.