Solve 1.03^x=440000/9800

$${ 1.03 }^{ x } = \frac{ 440000 }{ 9800 }$$

Solve for x

$$x=\log_{1.03}\left(\frac{2200}{49}\right)\approx 128.705903242$$

Steps Using Definition of Logarithm

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

$$1.03^{x}=\frac{2200}{49}$$

Take the logarithm of both sides of the equation.

$$\log(1.03^{x})=\log(\frac{2200}{49})$$

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

$$x\log(1.03)=\log(\frac{2200}{49})$$

Divide both sides by \log(1.03).

$$x=\frac{\log(\frac{2200}{49})}{\log(1.03)}$$

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

$$x=\log_{1.03}\left(\frac{2200}{49}\right)$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(1.03)}+\log_{1.03}\left(\frac{2200}{49}\right)$$

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