log1,5 (4/9)

asked by guest
on Sep 21, 2024 at 3:53 pm



MathBot Answer:

The sum of \(\log\left( 1 \right)\) and \(5 \cdot \frac{4}{9}\) is:

\[\begin{aligned}&=\left(\log\left( 1 \right)\right) + \left(5 \cdot \frac{4}{9}\right) \\\\ &= \log{\left(1 \right)} + 5 \cdot 4 \cdot \frac{1}{9} \\\\ &= \frac{20}{9} \\\\\end{aligned}\]

The difference of \(\log\left( 1 \right)\) and \(5 \cdot \frac{4}{9}\) is:

\[\begin{aligned}&=\left(\log\left( 1 \right)\right) - \left(5 \cdot \frac{4}{9}\right) \\\\ &= - \frac{20}{9} + \log{\left(1 \right)} \\\\ &= - \frac{20}{9} \\\\\end{aligned}\]

The product of \(\log\left( 1 \right)\) and \(5 \cdot \frac{4}{9}\) is:

\[\begin{aligned}&=\left(\log\left( 1 \right)\right) \cdot \left(5 \cdot \frac{4}{9}\right) \\\\ &= \frac{20 \log{\left(1 \right)}}{9} \\\\ &= 0 \\\\\end{aligned}\]

The quotient of \(\log\left( 1 \right)\) and \(5 \cdot \frac{4}{9}\) is:

\[\begin{aligned}&= \frac{\left(\log\left( 1 \right)\right)}{\left(5 \cdot \frac{4}{9}\right)} \\\\ &= \frac{9 \log{\left(1 \right)}}{20} \\\\ &= 0 \\\\\end{aligned}\]

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