Solve 525(1,048)^n-1/(1,048)frac{1{4}}=3125

$$525 \frac { ( 1,048 ) ^ { n } - 1 } { ( 1,048 ) \frac { 1 } { 4 } } = 3125$$

Solve for n

$$n=\log_{1048}\left(\frac{32771}{21}\right)\approx 1.057247826$$

Steps Using Definition of Logarithm

Divide both sides by 525.

$$\frac{1048^{n}-1}{1048\times \frac{1}{4}}=\frac{3125}{525}$$

Reduce the fraction \frac{3125}{525} to lowest terms by extracting and canceling out 25.

$$\frac{1048^{n}-1}{1048\times \frac{1}{4}}=\frac{125}{21}$$

Multiply 1048 and \frac{1}{4} to get 262.

$$\frac{1048^{n}-1}{262}=\frac{125}{21}$$

Divide each term of 1048^{n}-1 by 262 to get \frac{1}{262}\times 1048^{n}-\frac{1}{262}.

$$\frac{1}{262}\times 1048^{n}-\frac{1}{262}=\frac{125}{21}$$

Add \frac{1}{262} to both sides of the equation.

$$\frac{1}{262}\times 1048^{n}=\frac{32771}{5502}$$

Multiply both sides by 262.

$$1048^{n}=\frac{32771}{21}$$

Take the logarithm of both sides of the equation.

$$\log(1048^{n})=\log(\frac{32771}{21})$$

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

$$n\log(1048)=\log(\frac{32771}{21})$$

Divide both sides by \log(1048).

$$n=\frac{\log(\frac{32771}{21})}{\log(1048)}$$

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

$$n=\log_{1048}\left(\frac{32771}{21}\right)$$