Solve 14000=7000(1+0.96/4)^4x

$$14000 = 7000 ( 1 + \frac { 0.96 } { 4 } ) ^ { 4 x }$$

Solve for x

$$x=\frac{\log_{1.24}\left(2\right)}{4}\approx 0.805567774$$

Steps Using Definition of Logarithm

Divide both sides by 7000.

$$\frac{14000}{7000}=\left(1+\frac{0.96}{4}\right)^{4x}$$

Divide 14000 by 7000 to get 2.

$$2=\left(1+\frac{0.96}{4}\right)^{4x}$$

Expand \frac{0.96}{4} by multiplying both numerator and the denominator by 100.

$$2=\left(1+\frac{96}{400}\right)^{4x}$$

Reduce the fraction \frac{96}{400} to lowest terms by extracting and canceling out 16.

$$2=\left(1+\frac{6}{25}\right)^{4x}$$

Add 1 and \frac{6}{25} to get \frac{31}{25}.

$$2=\left(\frac{31}{25}\right)^{4x}$$

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

$$\left(\frac{31}{25}\right)^{4x}=2$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{31}{25}\right)^{4x})=\log(2)$$

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

$$4x\log(\frac{31}{25})=\log(2)$$

Divide both sides by \log(\frac{31}{25}).

$$4x=\frac{\log(2)}{\log(\frac{31}{25})}$$

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

$$4x=\log_{\frac{31}{25}}\left(2\right)$$

Divide both sides by 4.

$$x=\frac{\ln(2)}{4\ln(\frac{31}{25})}$$

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{2\ln(1.24)}+\frac{\log_{1.24}\left(2\right)}{4}$$

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