# How do you solve for x in log_6 3x-log_6 (x+1)=log_6 1?

Feb 14, 2015

You can use the property of the log:
${\log}_{a} M - {\log}_{a} N = {\log}_{a} \left(\frac{M}{N}\right)$
So you have:
${\log}_{6} \left(\frac{3 x}{x + 1}\right) = {\log}_{6} \left(1\right)$
${\log}_{6} \left(\frac{3 x}{x + 1}\right) = 0$
Because from the definition od logarithm:
${\log}_{a} b = x \implies {a}^{x} = b$
and: ${\log}_{6} \left(1\right) = 0$

as for:
${\log}_{6} \left(\frac{3 x}{x + 1}\right) = 0$
again you get:
$\left(\frac{3 x}{x + 1}\right) = {6}^{0} = 1$
$3 x = x + 1$
$x = \frac{1}{2}$