Hi Guys, new to the forum, does anyone know how to solve this question?
Solve the equation lg <x + 12> = 1 + (2 - mx)
hi if im not wrong...
lg <x + 12> = 1 + (2 - mx)
lg<x + 12> = 3 - mx
lg<x + 12> - lg<10^3> + lg<10^(mx)> = 0
lg<x + 12> + lg<10^(mx - 3)> = 0
lg<( x + 12 )( 10^(mx - 3) )> = 0
( x + 12 )( 10^(mx - 3) = 1
x + 12 = 1 and 10^(mx - 3) = 1 --(2)
x = -11 and sub into 2nd eqn
10^(-11m - 3) = 10^0
-11m = 3
m = - 3/4