Question

Solve for x:
log2x+log6=1+log12​

Answer 1

$log_{}(2x) + log_{}(6) = 1 + log_{}(12) \\ log_{}(2x) + log_{}(6) = log_{}(10) + log_{}(12) \\ log(2x) = log_{}(10) + log_{}(12) - log_{}(6) \\ log(2x) = 1 + 1.08 - 0.78 \\ log_{}(2x) = 1.3 = log_{}(19.952) \\ 2x = 19.952 \\ x = \frac{19.952}{2} \\ x = 9.976$