Solve 14,730=300(1(1+0,385/12)/frac{0.335{12}})^-n

$$14,730 = 300 ( \frac { 1 ( 1 + \frac { 0,385 } { 12 } ) } { \frac { 0.335 } { 12 } } ) ^ { - n }$$

Solve for n

$$n=\log_{\frac{67}{2400}}\left(49.1\right)\approx -1.088116493$$

Steps Using Definition of Logarithm

Divide both sides by 300.

$$\frac{14730}{300}=\left(\frac{1\left(1+\frac{0\times 385}{12}\right)}{\frac{0.335}{12}}\right)^{-n}$$

Reduce the fraction \frac{14730}{300} to lowest terms by extracting and canceling out 30.

$$\frac{491}{10}=\left(\frac{1\left(1+\frac{0\times 385}{12}\right)}{\frac{0.335}{12}}\right)^{-n}$$

Multiply 0 and 385 to get 0.

$$\frac{491}{10}=\left(\frac{1\left(1+\frac{0}{12}\right)}{\frac{0.335}{12}}\right)^{-n}$$

Zero divided by any non-zero number gives zero.

$$\frac{491}{10}=\left(\frac{1\left(1+0\right)}{\frac{0.335}{12}}\right)^{-n}$$

Add 1 and 0 to get 1.

$$\frac{491}{10}=\left(\frac{1\times 1}{\frac{0.335}{12}}\right)^{-n}$$

Multiply 1 and 1 to get 1.

$$\frac{491}{10}=\left(\frac{1}{\frac{0.335}{12}}\right)^{-n}$$

Expand \frac{0.335}{12} by multiplying both numerator and the denominator by 1000.

$$\frac{491}{10}=\left(\frac{1}{\frac{335}{12000}}\right)^{-n}$$

Reduce the fraction \frac{335}{12000} to lowest terms by extracting and canceling out 5.

$$\frac{491}{10}=\left(\frac{1}{\frac{67}{2400}}\right)^{-n}$$

Divide 1 by \frac{67}{2400} by multiplying 1 by the reciprocal of \frac{67}{2400}.

$$\frac{491}{10}=\left(1\times \frac{2400}{67}\right)^{-n}$$

Multiply 1 and \frac{2400}{67} to get \frac{2400}{67}.

$$\frac{491}{10}=\left(\frac{2400}{67}\right)^{-n}$$

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

$$\left(\frac{2400}{67}\right)^{-n}=\frac{491}{10}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{2400}{67}\right)^{-n})=\log(\frac{491}{10})$$

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

$$-n\log(\frac{2400}{67})=\log(\frac{491}{10})$$

Divide both sides by \log(\frac{2400}{67}).

$$-n=\frac{\log(\frac{491}{10})}{\log(\frac{2400}{67})}$$

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

$$-n=\log_{\frac{2400}{67}}\left(\frac{491}{10}\right)$$

Divide both sides by -1.

$$n=\frac{\ln(\frac{491}{10})}{-\ln(\frac{2400}{67})}$$