Solve {left(1+32/100right)}^x=200

$${ \left(1+ \frac{ 32 }{ 100 } \right) }^{ x } =200$$

Solve for x

$$x=\log_{\frac{33}{25}}\left(200\right)\approx 19.083975886$$

Steps Using Definition of Logarithm

Reduce the fraction \frac{32}{100} to lowest terms by extracting and canceling out 4.

$$\left(1+\frac{8}{25}\right)^{x}=200$$

Add 1 and \frac{8}{25} to get \frac{33}{25}.

$$\left(\frac{33}{25}\right)^{x}=200$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{33}{25}\right)^{x})=\log(200)$$

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

$$x\log(\frac{33}{25})=\log(200)$$

Divide both sides by \log(\frac{33}{25}).

$$x=\frac{\log(200)}{\log(\frac{33}{25})}$$

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

$$x=\log_{\frac{33}{25}}\left(200\right)$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(\frac{33}{25})}+\log_{\frac{33}{25}}\left(200\right)$$

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