Solve 26000=13000(1+.04125/12)^12t

$$26000 = 13000 ( 1 + \frac { .04125 } { 12 } ) ^ { 12 t }$$

Solve for t

$$t = \frac{\log_{1.0034375} {(2)}}{12} \approx 16.832432628$$

Steps Using Definition of Logarithm

Divide both sides by 13000.

$$\frac{26000}{13000}=\left(1+\frac{0.04125}{12}\right)^{12t}$$

Divide 26000 by 13000 to get 2.

$$2=\left(1+\frac{0.04125}{12}\right)^{12t}$$

Expand \frac{0.04125}{12} by multiplying both numerator and the denominator by 100000.

$$2=\left(1+\frac{4125}{1200000}\right)^{12t}$$

Reduce the fraction \frac{4125}{1200000} to lowest terms by extracting and canceling out 375.

$$2=\left(1+\frac{11}{3200}\right)^{12t}$$

Add 1 and \frac{11}{3200} to get \frac{3211}{3200}.

$$2=\left(\frac{3211}{3200}\right)^{12t}$$

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

$$\left(\frac{3211}{3200}\right)^{12t}=2$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{3211}{3200}\right)^{12t})=\log(2)$$

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

$$12t\log(\frac{3211}{3200})=\log(2)$$

Divide both sides by \log(\frac{3211}{3200}).

$$12t=\frac{\log(2)}{\log(\frac{3211}{3200})}$$

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

$$12t=\log_{\frac{3211}{3200}}\left(2\right)$$

Divide both sides by 12.

$$t=\frac{\ln(2)}{12\ln(\frac{3211}{3200})}$$