How do you solve $(\frac{1}{4}) e^{- 2 t} = 0.1$ $(1/4) e^(-2 t) = 0.1$ ?

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Answer 1 · 2016-02-04T15:29:18

$t = 0.458145$ $t=0.458145$

Start from the given equation

$(\frac{1}{4}) \cdot e^{- 2 t} = 0.1$ $(1/4)*e^(-2t)=0.1$

$4 \cdot (\frac{1}{4}) \cdot e^{- 2 t} = 4 (0.1)$ $4*(1/4)*e^(-2t)=4(0.1)$ Multiply both sides by $4$ $4$

$e^{- 2 t} = 0.4$ $e^(-2t)=0.4$

${ln e}^{- 2 t} = ln 0.4$ $ln e^(-2t)=ln 0.4$ Take the logarithm of both sides

$- 2 t = ln 0.4$ $-2t=ln 0.4$

$\frac{- 2 t}{- 2} = \frac{ln 0.4}{- 2}$ $(-2t)/-2=(ln 0.4)/-2$ Divide both sides by $- 2$ $-2$

$(cancel(-2)t)/cancel(-2)=(ln 0.4)/-2$

$t=(ln 0.4)/-2$

$t=0.458145$

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How do you solve (1/4) e^(-2 t) = 0.1(14)e−2t=0.1(1/4) e^(-2 t) = 0.1?