Solve 2.400/1200=(109/100)^x

$$\frac { 2.400 } { 1200 } = ( \frac { 109 } { 100 } ) ^ { x }$$

Solve for x

$$x=-\log_{1.09}\left(500\right)\approx -72.113880615$$

Steps Using Definition of Logarithm

Expand \frac{2.4}{1200} by multiplying both numerator and the denominator by 10.

$$\frac{24}{12000}=\left(\frac{109}{100}\right)^{x}$$

Reduce the fraction \frac{24}{12000} to lowest terms by extracting and canceling out 24.

$$\frac{1}{500}=\left(\frac{109}{100}\right)^{x}$$

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

$$\left(\frac{109}{100}\right)^{x}=\frac{1}{500}$$

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

$$\left(\frac{109}{100}\right)^{x}=0.002$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{109}{100}\right)^{x})=\log(0.002)$$

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

$$x\log(\frac{109}{100})=\log(0.002)$$

Divide both sides by \log(\frac{109}{100}).

$$x=\frac{\log(0.002)}{\log(\frac{109}{100})}$$

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

$$x=\log_{\frac{109}{100}}\left(0.002\right)$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(1.09)}-\log_{1.09}\left(500\right)$$

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