Solve 700=135{(e)^-0.00866x}

$$700=135 {(e)^{ -0.00866x }}$$

Solve for x

$$x=\frac{150000\ln(3)-50000\ln(140)}{433}\approx -190.046831017$$

Steps Using Definition of Logarithm

Divide both sides by 135.

$$\frac{700}{135}=e^{-0.00866x}$$

Reduce the fraction \frac{700}{135} to lowest terms by extracting and canceling out 5.

$$\frac{140}{27}=e^{-0.00866x}$$

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

$$e^{-0.00866x}=\frac{140}{27}$$

Take the logarithm of both sides of the equation.

$$\log(e^{-0.00866x})=\log(\frac{140}{27})$$

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

$$-0.00866x\log(e)=\log(\frac{140}{27})$$

Divide both sides by \log(e).

$$-0.00866x=\frac{\log(\frac{140}{27})}{\log(e)}$$

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

$$-0.00866x=\log_{e}\left(\frac{140}{27}\right)$$

Divide both sides of the equation by -0.00866, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=\frac{\ln(\frac{140}{27})}{-0.00866}$$

Solve for x (complex solution)

$$x=-\frac{i\times 100000\pi n_{1}}{433}+\frac{150000\ln(3)}{433}-\frac{50000\ln(140)}{433}$$

$$n_{1}\in \mathrm{Z}$$