Solve 25000*(1+0.03/365)^x=1000000

$$25000 \times ( 1 + \frac { 0.03 } { 365 } ) ^ { x } = 1000000$$

Solve for x

$$x=\log_{\frac{36503}{36500}}\left(40\right)\approx 44883.211106122$$

Steps Using Definition of Logarithm

Divide both sides by 25000.

$$\left(1+\frac{0.03}{365}\right)^{x}=\frac{1000000}{25000}$$

Divide 1000000 by 25000 to get 40.

$$\left(1+\frac{0.03}{365}\right)^{x}=40$$

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

$$\left(1+\frac{3}{36500}\right)^{x}=40$$

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

$$\left(\frac{36503}{36500}\right)^{x}=40$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{36503}{36500}\right)^{x})=\log(40)$$

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

$$x\log(\frac{36503}{36500})=\log(40)$$

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

$$x=\frac{\log(40)}{\log(\frac{36503}{36500})}$$

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

$$x=\log_{\frac{36503}{36500}}\left(40\right)$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(\frac{36503}{36500})}+\log_{\frac{36503}{36500}}\left(40\right)$$

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