Solve {left(1+3.2/100right)}^x=200

$${ \left(1+ \frac{ 3.2 }{ 100 } \right) }^{ x } =200$$

Solve for x

$$x=\log_{1.032}\left(200\right)\approx 168.207669123$$

Steps Using Definition of Logarithm

Expand \frac{3.2}{100} by multiplying both numerator and the denominator by 10.

$$\left(1+\frac{32}{1000}\right)^{x}=200$$

Reduce the fraction \frac{32}{1000} to lowest terms by extracting and canceling out 8.

$$\left(1+\frac{4}{125}\right)^{x}=200$$

Add 1 and \frac{4}{125} to get \frac{129}{125}.

$$\left(\frac{129}{125}\right)^{x}=200$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{129}{125}\right)^{x})=\log(200)$$

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

$$x\log(\frac{129}{125})=\log(200)$$

Divide both sides by \log(\frac{129}{125}).

$$x=\frac{\log(200)}{\log(\frac{129}{125})}$$

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

$$x=\log_{\frac{129}{125}}\left(200\right)$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(1.032)}+\log_{1.032}\left(200\right)$$

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