# How do you solve log_3 +log_3 (2x)=log_3 56?

Aug 10, 2016

The operand is missing in the first term. If it is a, the answer is $\frac{28}{a}$.

#### Explanation:

Assuming that the missing operand in the first term is a,

the equation is

${\log}_{3} a + \log 3 \left(2 x\right) = {\log}_{3} 56$

So, ${\log}_{3} \left(a X 2 x\right) = {\log}_{3} 56.$

And so, $2 a x = 56 \mathmr{and} x = \frac{28}{a}$.