e^((ln(8)-(1/96))72)

asked by guest
on Apr 03, 2025 at 2:24 am



You asked:

Evaluate the expression: e(ln⁡(8)−196)⋅72{e}^{\left(\ln\left( 8 \right) - \frac{1}{96}\right) \cdot 72}

MathBot Answer:

e(ln⁡(8)−196)⋅72=105312291668557186697918027683670432318895095400549111254310977536e34≈49746004176732642064128247766053742286521546510991065993704027520.68597820169346900685820113347923{e}^{\left(\ln\left( 8 \right) - \frac{1}{96}\right) \cdot 72} = \frac{105312291668557186697918027683670432318895095400549111254310977536}{e^{\frac{3}{4}}} \approx 49746004176732642064128247766053742286521546510991065993704027520.68597820169346900685820113347923

e(ln⁡(8)−196)⋅72=e(log⁡(8)−196)⋅72=e(−196+log⁡(8))⋅72=e−34+72⋅log⁡(8)=105312291668557186697918027683670432318895095400549111254310977536⋅e−34\begin{aligned}{e}^{\left(\ln\left( 8 \right) - \frac{1}{96}\right) \cdot 72}&=e^{\left(\log{\left(8 \right)} - \frac{1}{96}\right) \cdot 72}\\&=e^{\left(- \frac{1}{96} + \log{\left(8 \right)}\right) \cdot 72}\\&=e^{- \frac{3}{4} + 72 \cdot \log{\left(8 \right)}}\\&=105312291668557186697918027683670432318895095400549111254310977536 \cdot e^{- \frac{3}{4}}\end{aligned}

ee is Euler's number, a mathematical constant that is approximately 2.718282.71828.

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