What is e^{-1/5 ln(-5M+2)} ?

The solution to e^{-1/5 ln(-5M+2)} is 1/(e^{\frac{ln(-5M+2)){5}}}

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simplify e^{-1/5 ln(-5M+2)}

simplify

e^{- \frac{1}{5} l n (- 5 M + 2)}

\frac{1}{e ^{\frac{l n ( - 5 M + 2 )}{5}}}

Solution steps

e^{- \frac{1}{5} l n (- 5 M + 2)}

Apply exponent rule:

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

= \frac{1}{e ^{\frac{1}{5} l n (- 5 M + 2)}}

e^{\frac{1}{5} l n (- 5 M + 2)} = e^{\frac{l n ( - 5 M + 2 )}{5}}

= \frac{1}{e ^{\frac{l n ( - 5 M + 2 )}{5}}}

2 (e^{- y^{2}})

(4.7 x 1 0^{3}) (1.1 x 1 0^{6})

\frac{x - 5}{12 - x}

(3 e)^{3} (25 e)^{2} 36 e^{36}

0.8418 \cdot 1 0^{- 15}

What is e^{-1/5 ln(-5M+2)} ?
The solution to e^{-1/5 ln(-5M+2)} is 1/(e^{\frac{ln(-5M+2)){5}}}