Solve 250000*(1+0.03/365)^x=10000

$$250000 \times ( 1 + \frac { 0.03 } { 365 } ) ^ { x } = 10000$$

Solve for x

$$x=-2\log_{\frac{36503}{36500}}\left(5\right)\approx -39164.598618377$$

Steps Using Definition of Logarithm

Divide both sides by 250000.

$$\left(1+\frac{0.03}{365}\right)^{x}=\frac{10000}{250000}$$

Reduce the fraction \frac{10000}{250000} to lowest terms by extracting and canceling out 10000.

$$\left(1+\frac{0.03}{365}\right)^{x}=\frac{1}{25}$$

Expand \frac{0.03}{365} by multiplying both numerator and the denominator by 100.

$$\left(1+\frac{3}{36500}\right)^{x}=\frac{1}{25}$$

Add 1 and \frac{3}{36500} to get \frac{36503}{36500}.

$$\left(\frac{36503}{36500}\right)^{x}=\frac{1}{25}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{36503}{36500}\right)^{x})=\log(\frac{1}{25})$$

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

$$x\log(\frac{36503}{36500})=\log(\frac{1}{25})$$

Divide both sides by \log(\frac{36503}{36500}).

$$x=\frac{\log(\frac{1}{25})}{\log(\frac{36503}{36500})}$$

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

$$x=\log_{\frac{36503}{36500}}\left(\frac{1}{25}\right)$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(\frac{36503}{36500})}-2\log_{\frac{36503}{36500}}\left(5\right)$$

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