Solve 0.4=1.6*e^-D(1-0.67)

$$0.4 = 1.6 \times e ^ { - D ( 1 - 0.67 ) }$$

Solve for D

$$D=\frac{200\ln(2)}{33}\approx 4.200892003$$

Steps Using Definition of Logarithm

Divide both sides by 1.6.

$$\frac{0.4}{1.6}=e^{\left(-D\right)\left(1-0.67\right)}$$

Expand \frac{0.4}{1.6} by multiplying both numerator and the denominator by 10.

$$\frac{4}{16}=e^{\left(-D\right)\left(1-0.67\right)}$$

Reduce the fraction \frac{4}{16} to lowest terms by extracting and canceling out 4.

$$\frac{1}{4}=e^{\left(-D\right)\left(1-0.67\right)}$$

Subtract 0.67 from 1 to get 0.33.

$$\frac{1}{4}=e^{\left(-D\right)\times 0.33}$$

Swap sides so that all variable terms are on the left hand side.

$$e^{\left(-D\right)\times 0.33}=\frac{1}{4}$$

Multiply -1 and 0.33 to get -0.33.

$$e^{-0.33D}=\frac{1}{4}$$

Use the rules of exponents and logarithms to solve the equation.

$$e^{-0.33D}=0.25$$

Take the logarithm of both sides of the equation.

$$\log(e^{-0.33D})=\log(0.25)$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$-0.33D\log(e)=\log(0.25)$$

Divide both sides by \log(e).

$$-0.33D=\frac{\log(0.25)}{\log(e)}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$-0.33D=\log_{e}\left(0.25\right)$$

Divide both sides of the equation by -0.33, which is the same as multiplying both sides by the reciprocal of the fraction.

$$D=-\frac{2\ln(2)}{-0.33}$$