Question

Resolva, em Reais, a seguinte equação logarítmica:

log (1+2^x) + x = x . log 5 + log 6

Answer 1

$log~(1+2^x) +x = x~log~5+log~6\\ \\ log~(1+2^x) = x~log(5\cdot6) - x\\ \\ log~(1+2^x) = x~log~30-x\\ \\ log~(1+2^x) = x(log~30-1)\\ \\ log~(1+2^x) = x(log~30-log~10)\\ \\ log~(1+2^x) = x(log~\frac{30}{10})\\ \\ log~(1+2^x) = x~log~3\\ \\ log~(1+2^x) = log~3^x\\ \\ 1+2^x = 3^x\\ \\ \boxed{x=1}$