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Lerok [7]
4 years ago
14

Identify the error in the student solution shown below. Find the correct answer.

Mathematics
2 answers:
marshall27 [118]4 years ago
8 0
\bf 2ln(x)=ln(3x)-[ln(9)-2ln(3)]
\\\\\\
ln(x^2)=ln(3x)-[ln(9)-ln(3^2)]
\\\\\\
ln(x^2)=ln(3x)-[ln(9)-ln(9)]
\\\\\\
ln(x^2)=ln(3x)-\stackrel{\stackrel{\textit{it happened right here}}{\downarrow }}{\left[ \cfrac{ln(9)}{ln(9)} \right]}
\\\\\\
ln(x^2)=ln(3x)-[1]\impliedby \textit{recall }\frac{same}{same}=1\ne 0
lara [203]4 years ago
3 0

The answer is Since 0 in ln(3x) - 0 is not a logarithm, the property of logarithms cannot be used here.

The difference shown cannot be written as a quotient of logarithms.

The step ln(x2) = ln(3x) - (0) reduces to

ln(x2) = ln(3x).

The possible solutions are 0 and 3, with 0 being extraneous.

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