How do you simplify 2log 3+ log4-log6?

Feb 13, 2016

$\log \left(6\right)$

Explanation:

First, simplify $2 \log \left(3\right)$ through the rule:

$b \cdot \log \left(a\right) = \log \left({a}^{b}\right)$

Thus, $2 \log \left(3\right) = \log \left({3}^{2}\right) = \log \left(9\right)$.

Substitute this back into the original expression:

$= \log \left(9\right) + \log \left(4\right) - \log \left(6\right)$

We can now use the following rule regarding the addition of logarithms (with the same base, which we do have in this scenario):

$\log \left(a\right) + \log \left(b\right) = \log \left(a b\right)$

Thus the first two logarithms can be combined as follows:

$\log \left(9\right) + \log \left(4\right) = \log \left(9 \cdot 4\right) = \log \left(36\right)$

Substituting this back into the expression, we obtain:

$= \log \left(36\right) - \log \left(6\right)$

Now, to subtract logarithms with the same base, we use the rule:

$\log \left(a\right) - \log \left(b\right) = \log \left(\frac{a}{b}\right)$

Thus, the expression becomes

=log(36/6)=color(blue)(log(6)