Alright, let's solve the logarithmic equation step-by-step:
Given the equation:
[tex]\[ \log_4(x) + \log_4(x - 15) = 2 \][/tex]
We can use the property of logarithms that states:
[tex]\[ \log_b(m) + \log_b(n) = \log_b(m \cdot n) \][/tex]
Applying this property to our equation, we get:
[tex]\[ \log_4(x) + \log_4(x - 15) = \log_4(x(x - 15)) \][/tex]
So, the equation becomes:
[tex]\[ \log_4(x(x - 15)) = 2 \][/tex]
Since we have this logarithmic expression equal to 2, we can rewrite it in its exponential form using the definition of logarithms:
[tex]\[ \log_b(A) = C \implies b^C = A \][/tex]
Therefore:
[tex]\[ 4^2 = x(x - 15) \][/tex]
Simplify:
[tex]\[ 16 = x(x - 15) \][/tex]
Expand the right side:
[tex]\[ 16 = x^2 - 15x \][/tex]
Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 - 15x - 16 = 0 \][/tex]
Now, we need to solve this quadratic equation. We can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], the coefficients are:
[tex]\[ a = 1, \quad b = -15, \quad c = -16 \][/tex]
Plugging these values into the quadratic formula, we get:
[tex]\[ x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(1)(-16)}}{2(1)} \][/tex]
[tex]\[ x = \frac{15 \pm \sqrt{225 + 64}}{2} \][/tex]
[tex]\[ x = \frac{15 \pm \sqrt{289}}{2} \][/tex]
[tex]\[ x = \frac{15 \pm 17}{2} \][/tex]
This gives us two potential solutions:
[tex]\[ x_1 = \frac{15 + 17}{2} = \frac{32}{2} = 16 \][/tex]
[tex]\[ x_2 = \frac{15 - 17}{2} = \frac{-2}{2} = -1 \][/tex]
Since [tex]\( x = -1 \)[/tex] is not a valid solution due to the constraints of logarithmic functions (logarithms are not defined for non-positive arguments), we discard [tex]\( x = -1 \)[/tex].
Thus, the only valid solution is:
[tex]\[ x = 16 \][/tex]
So:
[tex]\[ \boxed{16} \][/tex]
