Question 965036
the first one is solved as follows:


log(2-x) = .5


this is true if and only 10^.5 = 2-x


square both sides of this equation to get:


10 = (2-x)^2


simplify to get:


10 = 4 - 4x + x^2


subtract 10 from both sides and re-arrange the terms to get:


x^2 - 4x - 6 = 0


factor using the quadratic formula to get:


x = 5.16227766


x = -1.16227766


both values work.


just replace them in the original equation and you'll see.


the second problem is solved as follows:


log(x + 25) = log(x + 10) + log 4  


this is equivalent to log(x + 25) = log(4 * (x + 10) which is equivalent to log(x + 25) = log(4x + 40).


this is true if and only if x + 25 = 4x + 40


solve for x to get x = -5.


replace x with -5 in the original equation and you'll see that they're equivalent.


log(-5 + 25) = log(20).\


log(x + 10) + log(4) is equal to log(5) + log(4) which is equal to log(5 * 4) which is equal to log(20).


the rule of logs that are used are:


logb(x) = y if and only if b^y = x


logb(a) + logb(b) = logb(a * b)