What is 1-e^{-θ((yθ)/(-ln(2)-ln(θ)))} ?

The solution to 1-e^{-θ((yθ)/(-ln(2)-ln(θ)))} is 1-e^{(θ^2y)/(ln(2θ))}

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simplify 1-e^{-θ((yθ)/(-ln(2)-ln(θ)))}

simplify

1 - e^{- θ (\frac{y θ}{- l n ( 2 ) - l n ( θ )})}

1 - e^{\frac{θ ^{2} y}{l n ( 2 θ )}}

Solution steps

1 - e^{- θ (\frac{y θ}{- l n ( 2 ) - l n ( θ )})}

Rewrite

- ln (2) - ln (θ)

- (ln (2) + ln (θ))

= 1 - e^{- (\frac{θ y}{- ( l n ( θ ) + l n ( 2 ) )}) θ}

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

ln (2) + ln (θ) = ln (2 θ)

= 1 - e^{- θ \frac{y θ}{- l n ( 2 θ )}}

e^{- θ \frac{y θ}{- l n ( 2 θ )}} = e^{\frac{θ ^{2} y}{l n ( 2 θ )}}

= 1 - e^{\frac{θ ^{2} y}{l n ( 2 θ )}}

- \frac{1}{2} ln (1 - 2 v) - ln (x)

2 ln (2 m)

lo g_{10} (z^{4} y \cdot x^{9})

2550 \cdot ln (\frac{2950000}{2950000 - 2200000})

ln (\frac{500}{250})

What is 1-e^{-θ((yθ)/(-ln(2)-ln(θ)))} ?
The solution to 1-e^{-θ((yθ)/(-ln(2)-ln(θ)))} is 1-e^{(θ^2y)/(ln(2θ))}