-(1/8.55E-4)ln[-ln(0.3046)]=

asked by guest
on Nov 26, 2024 at 12:35 pm



You asked:

Evaluate the expression: \(-\left( \left(\frac{1}{8.55 E} - 4\right) \cdot \ln\left( -\ln\left( 0.3046 \right) \right) \right)\)

MathBot Answer:

Evaluated



\(\displaystyle -\left( \left(\frac{1}{8.55 E} - 4\right) \cdot \ln\left( -\ln\left( 0.3046 \right) \right) \right) = - \left(-4 + \frac{20}{171 E}\right) \log{\left(- \log{\left(\frac{1523}{5000} \right)} \right)} \)

Expanded

\[-\left( \left(\frac{1}{8.55 E} - 4\right) \cdot \ln\left( -\ln\left( 0.3046 \right) \right) \right) = 4 \log{\left(- \log{\left(3046 \right)} + 4 \log{\left(10 \right)} \right)} - \frac{20 \log{\left(- \log{\left(3046 \right)} + 4 \log{\left(10 \right)} \right)}}{171 E}\]

Factored

\[-\left( \left(\frac{1}{8.55 E} - 4\right) \cdot \ln\left( -\ln\left( 0.3046 \right) \right) \right) = \frac{4 \cdot \left(171 E - 5\right) \log{\left(- \log{\left(3046 \right)} + 4 \log{\left(10 \right)} \right)}}{171 E}\]

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