To find the x-intercepts of the function f(x) = log base 3 of (x-1) + log base 3 of (2x+3), set the function equal to zero, combine the logarithms using logarithmic rules, and solve the resulting quadratic equation. The solutions are x = 1 and x = -3/2. Here option B is correct.
Set the function equal to zero. This is because we want to find the values of x that make the function have a value of zero, or in other words, the x-intercepts of the graph of the function. So, we have the equation:
log3(x-1) + log3(2x+3) = 0
Combine the logarithms. Since the function has two logarithms with the same base (3), we can use the rule of logarithms that states: log_a(b) + log_a(c) = log_a(b*c). Therefore, we can combine the two logarithms like this:
log3((x-1)(2x+3)) = 0
Set the argument of the logarithm equal to 1. Since the logarithm is equal to zero, its argument must be equal to the base raised to the power of zero, which is 1. So, we have the equation:
(x-1)(2x+3) = 1
Solve the quadratic equation. This equation is now a quadratic equation in terms of x. You can solve it using various methods, such as factoring, the quadratic formula, or completing the square. In this case, factoring gives us:
(x-1)(2x+3) = 0
(x-1) = 0 or (2x+3) = 0
x = 1 or x = -3/2
Therefore, the zeros of the function f(x) = log3(x-1) + log3(2x+3) are x = 1 and x = -3/2. Here option B is correct.
Complete question:
Find the zero(es) of the function f(x) = log_3(x - 1) + log_3(2x + 3).