Question

solve each equation. leave your answer in exact form.\\(\log_{2}(x + 9) + \log_{2}(x) = \log_{2}(4)\\)

Explanation:

Step1: Apply log addition rule

Using the property \(\log_a(M) + \log_a(N)=\log_a(MN)\), we combine the left - hand side:
\(\log_2((x + 9)x)=\log_2(4)\)

Step2: Use log equality property

If \(\log_a(b)=\log_a(c)\), then \(b = c\) (for \(a>0,a
eq1,b>0,c>0\)). So we have:
\(x(x + 9)=4\)
Expand the left - hand side: \(x^{2}+9x - 4=0\)

Step3: Solve quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 1\), \(b = 9\), \(c=-4\)), the quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)
\(x=\frac{-9\pm\sqrt{9^{2}-4\times1\times(-4)}}{2\times1}=\frac{-9\pm\sqrt{81 + 16}}{2}=\frac{-9\pm\sqrt{97}}{2}\)

Step4: Check domain

The domain of \(\log_2(x+9)\) and \(\log_2(x)\) requires \(x+9>0\) and \(x>0\), so \(x>0\)
\(\frac{-9-\sqrt{97}}{2}<0\) (discard), \(\frac{-9 + \sqrt{97}}{2}\approx\frac{-9+9.849}{2}=\frac{0.849}{2}>0\) (valid)

Answer:

\(x=\frac{-9 + \sqrt{97}}{2}\)