Solve 19100=6000*1.045^n-1/1.045-1

$$19100 = 6000 \times \frac{ { 1.045 }^{ n } -1 }{ 1.045-1 }$$

Solve for n

$$n=\log_{1.045}\left(1.14325\right)\approx 3.041448351$$

Steps Using Definition of Logarithm

Divide both sides by 6000.

$$\frac{19100}{6000}=\frac{1.045^{n}-1}{1.045-1}$$

Reduce the fraction \frac{19100}{6000} to lowest terms by extracting and canceling out 100.

$$\frac{191}{60}=\frac{1.045^{n}-1}{1.045-1}$$

Subtract 1 from 1.045 to get 0.045.

$$\frac{191}{60}=\frac{1.045^{n}-1}{0.045}$$

Divide each term of 1.045^{n}-1 by 0.045 to get \frac{1.045^{n}}{0.045}+\frac{-1}{0.045}.

$$\frac{191}{60}=\frac{1.045^{n}}{0.045}+\frac{-1}{0.045}$$

Expand \frac{-1}{0.045} by multiplying both numerator and the denominator by 1000.

$$\frac{191}{60}=\frac{1.045^{n}}{0.045}+\frac{-1000}{45}$$

Reduce the fraction \frac{-1000}{45} to lowest terms by extracting and canceling out 5.

$$\frac{191}{60}=\frac{1.045^{n}}{0.045}-\frac{200}{9}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1.045^{n}}{0.045}-\frac{200}{9}=\frac{191}{60}$$

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

$$\frac{200}{9}\times 1.045^{n}-\frac{200}{9}=\frac{191}{60}$$

Add \frac{200}{9} to both sides of the equation.

$$\frac{200}{9}\times 1.045^{n}=\frac{4573}{180}$$

Divide both sides of the equation by \frac{200}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$1.045^{n}=\frac{4573}{4000}$$

Take the logarithm of both sides of the equation.

$$\log(1.045^{n})=\log(\frac{4573}{4000})$$

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

$$n\log(1.045)=\log(\frac{4573}{4000})$$

Divide both sides by \log(1.045).

$$n=\frac{\log(\frac{4573}{4000})}{\log(1.045)}$$

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

$$n=\log_{1.045}\left(\frac{4573}{4000}\right)$$