Solve 2000=1000(1+0.05/365)^365x

$$2000 = 1000 ( 1 + \frac { 0.05 } { 365 } ) ^ { 365 x }$$

Solve for x

$$x = \frac{\log_{\frac{7301}{7300}} {(2)}}{365} \approx 13.863893106$$

Steps Using Definition of Logarithm

Divide both sides by 1000.

$$\frac{2000}{1000}=\left(1+\frac{0.05}{365}\right)^{365x}$$

Divide 2000 by 1000 to get 2.

$$2=\left(1+\frac{0.05}{365}\right)^{365x}$$

Expand \frac{0.05}{365} by multiplying both numerator and the denominator by 100.

$$2=\left(1+\frac{5}{36500}\right)^{365x}$$

Reduce the fraction \frac{5}{36500} to lowest terms by extracting and canceling out 5.

$$2=\left(1+\frac{1}{7300}\right)^{365x}$$

Add 1 and \frac{1}{7300} to get \frac{7301}{7300}.

$$2=\left(\frac{7301}{7300}\right)^{365x}$$

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

$$\left(\frac{7301}{7300}\right)^{365x}=2$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{7301}{7300}\right)^{365x})=\log(2)$$

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

$$365x\log(\frac{7301}{7300})=\log(2)$$

Divide both sides by \log(\frac{7301}{7300}).

$$365x=\frac{\log(2)}{\log(\frac{7301}{7300})}$$

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

$$365x=\log_{\frac{7301}{7300}}\left(2\right)$$

Divide both sides by 365.

$$x=\frac{\ln(2)}{365\ln(\frac{7301}{7300})}$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{365\ln(\frac{7301}{7300})}+\frac{\log_{\frac{7301}{7300}}\left(2\right)}{365}$$

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