Solve 250[(1+0.12/12)^12n-1/frac{0.12{12}}]=20417.42

$$250 [ \frac { ( 1 + \frac { 0.12 } { 12 } ) ^ { 12 n } - 1 } { \frac { 0.12 } { 12 } } ] = 20417.42$$

Solve for n

$$n = \frac{\log_{1.01} {(1.8166968)}}{12} \approx 5.000000468$$

Steps Using Definition of Logarithm

Divide both sides by 250.

$$\frac{\left(1+\frac{0.12}{12}\right)^{12n}-1}{\frac{0.12}{12}}=\frac{20417.42}{250}$$

Expand \frac{20417.42}{250} by multiplying both numerator and the denominator by 100.

$$\frac{\left(1+\frac{0.12}{12}\right)^{12n}-1}{\frac{0.12}{12}}=\frac{2041742}{25000}$$

Reduce the fraction \frac{2041742}{25000} to lowest terms by extracting and canceling out 2.

$$\frac{\left(1+\frac{0.12}{12}\right)^{12n}-1}{\frac{0.12}{12}}=\frac{1020871}{12500}$$

Expand \frac{0.12}{12} by multiplying both numerator and the denominator by 100.

$$\frac{\left(1+\frac{12}{1200}\right)^{12n}-1}{\frac{0.12}{12}}=\frac{1020871}{12500}$$

Reduce the fraction \frac{12}{1200} to lowest terms by extracting and canceling out 12.

$$\frac{\left(1+\frac{1}{100}\right)^{12n}-1}{\frac{0.12}{12}}=\frac{1020871}{12500}$$

Add 1 and \frac{1}{100} to get \frac{101}{100}.

$$\frac{\left(\frac{101}{100}\right)^{12n}-1}{\frac{0.12}{12}}=\frac{1020871}{12500}$$

Expand \frac{0.12}{12} by multiplying both numerator and the denominator by 100.

$$\frac{\left(\frac{101}{100}\right)^{12n}-1}{\frac{12}{1200}}=\frac{1020871}{12500}$$

Reduce the fraction \frac{12}{1200} to lowest terms by extracting and canceling out 12.

$$\frac{\left(\frac{101}{100}\right)^{12n}-1}{\frac{1}{100}}=\frac{1020871}{12500}$$

Divide each term of \left(\frac{101}{100}\right)^{12n}-1 by \frac{1}{100} to get \frac{\left(\frac{101}{100}\right)^{12n}}{\frac{1}{100}}+\frac{-1}{\frac{1}{100}}.

$$\frac{\left(\frac{101}{100}\right)^{12n}}{\frac{1}{100}}+\frac{-1}{\frac{1}{100}}=\frac{1020871}{12500}$$

Divide -1 by \frac{1}{100} by multiplying -1 by the reciprocal of \frac{1}{100}.

$$\frac{\left(\frac{101}{100}\right)^{12n}}{\frac{1}{100}}-100=\frac{1020871}{12500}$$

Subtract \frac{1020871}{12500} from both sides.

$$\frac{\left(\frac{101}{100}\right)^{12n}}{\frac{1}{100}}-100-\frac{1020871}{12500}=0$$

Subtract \frac{1020871}{12500} from -100 to get -\frac{2270871}{12500}.

$$\frac{\left(\frac{101}{100}\right)^{12n}}{\frac{1}{100}}-\frac{2270871}{12500}=0$$

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

$$100\times \left(\frac{101}{100}\right)^{12n}-\frac{2270871}{12500}=0$$

Add \frac{2270871}{12500} to both sides of the equation.

$$100\times \left(\frac{101}{100}\right)^{12n}=\frac{2270871}{12500}$$

Divide both sides by 100.

$$\left(\frac{101}{100}\right)^{12n}=\frac{2270871}{1250000}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{101}{100}\right)^{12n})=\log(\frac{2270871}{1250000})$$

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

$$12n\log(\frac{101}{100})=\log(\frac{2270871}{1250000})$$

Divide both sides by \log(\frac{101}{100}).

$$12n=\frac{\log(\frac{2270871}{1250000})}{\log(\frac{101}{100})}$$

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

$$12n=\log_{\frac{101}{100}}\left(\frac{2270871}{1250000}\right)$$

Divide both sides by 12.

$$n=\frac{\ln(\frac{2270871}{1250000})}{12\ln(\frac{101}{100})}$$

Solve for n (complex solution)

$$n=\frac{i\pi n_{1}}{6\ln(1.01)}-\frac{\log_{1.01}\left(\frac{1250000}{2270871}\right)}{12}$$

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