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Question
solve for x.
\log_{5}(x - 8) = \log_{5}(x - 4) + 1
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x = \square
Step1: Use logarithm property
Recall that \(1 = \log_{5}5\) (since \(\log_{a}a = 1\) for any \(a>0,a
eq1\)). So rewrite the equation:
\(\log_{5}(x - 8)=\log_{5}(x - 4)+\log_{5}5\)
Step2: Apply log addition rule
Use the property \(\log_{a}M+\log_{a}N=\log_{a}(M\times N)\) (for \(a > 0,a
eq1,M>0,N>0\)). Then the right - hand side becomes \(\log_{5}[5(x - 4)]\). So the equation is:
\(\log_{5}(x - 8)=\log_{5}(5x-20)\)
Step3: Set arguments equal
If \(\log_{a}M=\log_{a}N\), then \(M = N\) (for \(a>0,a
eq1,M>0,N>0\)). So we have:
\(x - 8=5x-20\)
Step4: Solve for x
Subtract \(x\) from both sides:
\(- 8 = 4x-20\)
Add 20 to both sides:
\(12 = 4x\)
Divide both sides by 4:
\(x = 3\)
Step5: Check domain
For \(\log_{5}(x - 8)\) to be defined, \(x-8>0\Rightarrow x > 8\). For \(\log_{5}(x - 4)\) to be defined, \(x - 4>0\Rightarrow x>4\). But when \(x = 3\), \(x-8=3 - 8=-5<0\) and \(x - 4=3 - 4=-1<0\), so \(x = 3\) does not satisfy the domain of the logarithmic functions.
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