To solve this equation, we need to use properties of logarithms.
lg(x^2 + 19) - lg(x + 1) = 1lg((x^2 + 19)/(x + 1)) = 1
10^1 = (x^2 + 19)/(x + 1)
10 = (x^2 + 19)/(x + 1)
10(x + 1) = x^2 + 1910x + 10 = x^2 + 190 = x^2 - 10x - 9
x = (-(-10) ± √((-10)^2 - 4(1)(-9)))/(2(1))x = (10 ± √(100 + 36))/2x = (10 ± √136)/2x = (10 ± 2√34)/2x = 5 ± √34
Therefore, the solutions to the equation lg(x^2 + 19) - lg(x + 1) = 1 are x = 5 + √34 and x = 5 - √34.
To solve this equation, we need to use properties of logarithms.
Apply the property of logarithms that states the difference of two logarithms is equal to the logarithm of the division of their arguments.lg(x^2 + 19) - lg(x + 1) = 1
Rewrite the equation in exponential form.lg((x^2 + 19)/(x + 1)) = 1
10^1 = (x^2 + 19)/(x + 1)
Simplify the equation.10 = (x^2 + 19)/(x + 1)
Cross multiply and solve for x.10(x + 1) = x^2 + 19
Use the quadratic formula to solve for x.10x + 10 = x^2 + 19
0 = x^2 - 10x - 9
x = (-(-10) ± √((-10)^2 - 4(1)(-9)))/(2(1))
x = (10 ± √(100 + 36))/2
x = (10 ± √136)/2
x = (10 ± 2√34)/2
x = 5 ± √34
Therefore, the solutions to the equation lg(x^2 + 19) - lg(x + 1) = 1 are x = 5 + √34 and x = 5 - √34.