Solve 7(1+11%)^x=7.31

$$7 ( 1 + 11 \% ) ^ { x } = 7.31$$

Solve for x

$$x=\log_{1.11}\left(\frac{731}{700}\right)\approx 0.415227274$$

Steps Using Definition of Logarithm

Divide both sides by 7.

$$\left(1+\frac{11}{100}\right)^{x}=\frac{7.31}{7}$$

Expand \frac{7.31}{7} by multiplying both numerator and the denominator by 100.

$$\left(1+\frac{11}{100}\right)^{x}=\frac{731}{700}$$

Add 1 and \frac{11}{100} to get \frac{111}{100}.

$$\left(\frac{111}{100}\right)^{x}=\frac{731}{700}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{111}{100}\right)^{x})=\log(\frac{731}{700})$$

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

$$x\log(\frac{111}{100})=\log(\frac{731}{700})$$

Divide both sides by \log(\frac{111}{100}).

$$x=\frac{\log(\frac{731}{700})}{\log(\frac{111}{100})}$$

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

$$x=\log_{\frac{111}{100}}\left(\frac{731}{700}\right)$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(1.11)}+\log_{1.11}\left(\frac{731}{700}\right)$$

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