Solve 4000=13000(1+8%)^-n

$$4000 = 13000 ( 1 + 8 \% ) ^ { - n }$$

Solve for n

$$n=\log_{\frac{25}{27}}\left(\frac{4}{13}\right)\approx 15.314956489$$

Steps Using Definition of Logarithm

Divide both sides by 13000.

$$\frac{4000}{13000}=\left(1+\frac{8}{100}\right)^{-n}$$

Reduce the fraction \frac{4000}{13000} to lowest terms by extracting and canceling out 1000.

$$\frac{4}{13}=\left(1+\frac{8}{100}\right)^{-n}$$

Reduce the fraction \frac{8}{100} to lowest terms by extracting and canceling out 4.

$$\frac{4}{13}=\left(1+\frac{2}{25}\right)^{-n}$$

Add 1 and \frac{2}{25} to get \frac{27}{25}.

$$\frac{4}{13}=\left(\frac{27}{25}\right)^{-n}$$

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

$$\left(\frac{27}{25}\right)^{-n}=\frac{4}{13}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{27}{25}\right)^{-n})=\log(\frac{4}{13})$$

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

$$-n\log(\frac{27}{25})=\log(\frac{4}{13})$$

Divide both sides by \log(\frac{27}{25}).

$$-n=\frac{\log(\frac{4}{13})}{\log(\frac{27}{25})}$$

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

$$-n=\log_{\frac{27}{25}}\left(\frac{4}{13}\right)$$

Divide both sides by -1.

$$n=\frac{\ln(\frac{4}{13})}{-\ln(\frac{27}{25})}$$