Solve 1.02*(1.02^x-1)=1.1*0.02

$$1.02 \times ( { 1.02 }^{ x } -1)=1.1 \times 0.02$$

Solve for x

$$x=\log_{1.02}\left(\frac{521}{510}\right)\approx 1.077600245$$

Steps Using Definition of Logarithm

Use the distributive property to multiply 1.02 by 1.02^{x}-1.

$$1.02\times 1.02^{x}-1.02=1.1\times 0.02$$

Multiply 1.1 and 0.02 to get 0.022.

$$1.02\times 1.02^{x}-1.02=0.022$$

Add 1.02 to both sides of the equation.

$$1.02\times 1.02^{x}=1.042$$

Divide both sides of the equation by 1.02, which is the same as multiplying both sides by the reciprocal of the fraction.

$$1.02^{x}=\frac{521}{510}$$

Take the logarithm of both sides of the equation.

$$\log(1.02^{x})=\log(\frac{521}{510})$$

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

$$x\log(1.02)=\log(\frac{521}{510})$$

Divide both sides by \log(1.02).

$$x=\frac{\log(\frac{521}{510})}{\log(1.02)}$$

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

$$x=\log_{1.02}\left(\frac{521}{510}\right)$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(1.02)}+\log_{1.02}\left(\frac{521}{510}\right)$$

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