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

1 Answer
Aug 10, 2016

Answer:

The operand is missing in the first term. If it is a, the answer is #28/a#.

Explanation:

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

the equation is

#log_3 a + log 3 (2x) = log_3 56#

So, #log_3 (a X 2x)=log_3 56.#

And so, #2ax=56 and x =28/a#.