log5(x+3)+log5(x-2)=log5(14)
Log5(x+3)(x-2)=log5(14)
(X+3)(x-2)=14
X^2+x-6=14
X^2+x-20=0
(X+5)(x-4)=0
X=-5, x=4
-5 is not a true solution because you can't take the log of a negative number
X=4
To solve the equation log5(x+3) + log5(x-2) = log5(14), we can use the properties of logarithms.
First, let's simplify the equation using the property that log(a) + log(b) = log(a * b):
log5((x+3)(x-2)) = log5(14)
Now, equating the logarithmic expressions on both sides of the equation, we get:
(x+3)(x-2) = 14
Expanding the equation:
x^2 - 2x + 3x - 6 = 14
Simplifying further:
x^2 + x - 20 = 0
This is now a quadratic equation. To solve it, we can factor or use the quadratic formula. Let's factor it:
(x + 5)(x - 4) = 0
Setting each factor equal to zero:
x + 5 = 0 or x - 4 = 0
Solving for x in each case, we find:
x = -5 or x = 4
So, the solutions for the given equation are x = -5 and x = 4.