Question

a. round answer to 4 decimal places. \\(log_{13}(x^2 - 7x) = log_{13}(9 + x)\\)

Explanation:

Step1: Use Logarithm Property

Since \(\log_{a}M=\log_{a}N\) implies \(M = N\) (for \(a>0,a
eq1,M>0,N>0\)), we have:
\(x^{2}-7x=9 + x\)

Step2: Rearrange to Quadratic Equation

\(x^{2}-7x - x-9 = 0\)
\(x^{2}-8x - 9=0\)

Step3: Factor the Quadratic

\((x - 9)(x+ 1)=0\)

Step4: Solve for x

\(x-9 = 0\) or \(x + 1=0\)
So \(x = 9\) or \(x=-1\)

Step5: Check Domain

For \(\log_{13}(x^{2}-7x)\) and \(\log_{13}(9 + x)\) to be defined:

• For \(x=-1\): \(x^{2}-7x=1 + 7 = 8>0\), \(9 + x=8>0\)
• For \(x = 9\): \(x^{2}-7x=81-63 = 18>0\), \(9 + x=18>0\)

But we need to check the original equation. Wait, the original equation is \(\log_{13}(x^{2}-7x)
eq\log_{13}(9 + x)\), so we need to find when \(x^{2}-7x
eq9 + x\) except where the logs are undefined. Wait, maybe I misread. If the equation was \(\log_{13}(x^{2}-7x)=\log_{13}(9 + x)\), the solutions are \(x = 9\) and \(x=-1\). But if it's \(
eq\), we need to find all \(x\) in the domain of the logs except \(x = 9\) and \(x=-1\). But maybe the original problem was a typo and it's \(=\). Assuming it's \(=\):

We have solutions \(x = 9\) (which is exact) and \(x=-1\). If we consider decimal places, but \(x = 9\) is an integer. If we made a mistake and the equation was different, but based on the given, if we assume the equation is \(\log_{13}(x^{2}-7x)=\log_{13}(9 + x)\), the solutions are \(x = 9\) (or \(x=-1\)). If we need to round to 4 decimal places, \(x = 9.0000\) or \(x=-1.0000\)

Wait, maybe the original problem was to solve \(\log_{13}(x^{2}-7x)=\log_{13}(9 + x)\) and round. So the solutions are \(x = 9\) (which is \(9.0000\)) and \(x=-1\) (which is \(-1.0000\))

Answer:

If the equation is \(\log_{13}(x^{2}-7x)=\log_{13}(9 + x)\), the solutions are \(x = 9.0000\) or \(x=-1.0000\) (depending on the context, if we take the positive solution \(x = 9.0000\))