To solve the equation lg(x+1) + lg(x+4) = 1, we can combine the logarithms using the properties of logarithms.
lg(x+1) + lg(x+4) = lg((x+1)(x+4))
Now, we can rewrite the equation as:
lg((x+1)(x+4)) = 1
Using the definition of logarithms, we know that 10^1 = (x+1)(x+4).
Therefore, we have:10 = x^2 + 5x + 4
Rearranging the equation:x^2 + 5x - 6 = 0
Now, we can factor the quadratic equation:(x + 6)(x - 1) = 0
Setting each factor to zero:x + 6 = 0 or x - 1 = 0
x = -6 or x = 1
Therefore, the solutions to the equation lg(x+1) + lg(x+4) = 1 are x = -6 and x = 1.
To solve the equation lg(x+1) + lg(x+4) = 1, we can combine the logarithms using the properties of logarithms.
lg(x+1) + lg(x+4) = lg((x+1)(x+4))
Now, we can rewrite the equation as:
lg((x+1)(x+4)) = 1
Using the definition of logarithms, we know that 10^1 = (x+1)(x+4).
Therefore, we have:
10 = x^2 + 5x + 4
Rearranging the equation:
x^2 + 5x - 6 = 0
Now, we can factor the quadratic equation:
(x + 6)(x - 1) = 0
Setting each factor to zero:
x + 6 = 0 or x - 1 = 0
x = -6 or x = 1
Therefore, the solutions to the equation lg(x+1) + lg(x+4) = 1 are x = -6 and x = 1.