Question

1+log3(5+x)=2log3(1−x )
not just (log3) the 3 should be below, bit down the log

Answer 1

a bit down

Answer 2

\[1+\log_3(5+x) = 2\log_3(1-x) \rightarrow \log_33 +\log_3(5+x) = 2\log_3(1-x)\]

\[\log_33(5+x) = \log_3(1-x)^2\]

15+3x = (1-x)^2

Solve you get two solutions...
x = -2
x = 7.

The only one acceptable is x = -2 because of the domain :

5+x > 0 => x > - 5
-x > -1 => x < 1

-5<x<1

So x = -2