What is e^{-ln(10-0.8v_{1})} ?

The solution to e^{-ln(10-0.8v_{1})} is 1/(-0.8v_{1)+10}

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simplify e^{-ln(10-0.8v_{1})}

simplify

e^{- l n (10 - 0.8 v_{1})}

\frac{1}{- 0.8 v _{1} + 10}

Solution steps

e^{- l n (10 - 0.8 v_{1})}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{l n (10 - 0.8 v_{1})}}

e^{l n (10 - 0.8 v_{1})} = 10 - 0.8 v_{1}

= \frac{1}{10 - 0.8 v _{1}}

(3 x^{4}) (4 x^{4})

105 g h

1.42 \cdot 1 0^{- 4}

e^{- (t - τ)}

5000 (1 + \frac{0.05}{12})^{- 36}

What is e^{-ln(10-0.8v_{1})} ?
The solution to e^{-ln(10-0.8v_{1})} is 1/(-0.8v_{1)+10}