Solve 399300=300000(1+10/100)^x

$$399300 = 300000 ( 1 + \frac { 10 } { 100 } ) ^ { x }$$

Solve for x

$$x=3$$

Steps Using Definition of Logarithm

Divide both sides by 300000.

$$\frac{399300}{300000}=\left(1+\frac{10}{100}\right)^{x}$$

Reduce the fraction \frac{399300}{300000} to lowest terms by extracting and canceling out 300.

$$\frac{1331}{1000}=\left(1+\frac{10}{100}\right)^{x}$$

Reduce the fraction \frac{10}{100} to lowest terms by extracting and canceling out 10.

$$\frac{1331}{1000}=\left(1+\frac{1}{10}\right)^{x}$$

Add 1 and \frac{1}{10} to get \frac{11}{10}.

$$\frac{1331}{1000}=\left(\frac{11}{10}\right)^{x}$$

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

$$\left(\frac{11}{10}\right)^{x}=\frac{1331}{1000}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{11}{10}\right)^{x})=\log(\frac{1331}{1000})$$

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

$$x\log(\frac{11}{10})=\log(\frac{1331}{1000})$$

Divide both sides by \log(\frac{11}{10}).

$$x=\frac{\log(\frac{1331}{1000})}{\log(\frac{11}{10})}$$

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

$$x=\log_{\frac{11}{10}}\left(\frac{1331}{1000}\right)$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(\frac{11}{10})}+\log_{\frac{11}{10}}\left(\frac{1331}{1000}\right)$$

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