Solve 1410=4500(0.95)^2x+580

$$1410 = 4500 ( 0.95 ) ^ { 2 x } + 580$$

Solve for x

$$x = \frac{\log_{0.95} {(\frac{83}{450})}}{2} \approx 16.477855392$$

Steps Using Definition of Logarithm

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

$$4500\times 0.95^{2x}+580=1410$$

Subtract 580 from both sides of the equation.

$$4500\times 0.95^{2x}=830$$

Divide both sides by 4500.

$$0.95^{2x}=\frac{83}{450}$$

Take the logarithm of both sides of the equation.

$$\log(0.95^{2x})=\log(\frac{83}{450})$$

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

$$2x\log(0.95)=\log(\frac{83}{450})$$

Divide both sides by \log(0.95).

$$2x=\frac{\log(\frac{83}{450})}{\log(0.95)}$$

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

$$2x=\log_{0.95}\left(\frac{83}{450}\right)$$

Divide both sides by 2.

$$x=\frac{\ln(\frac{83}{450})}{2\ln(\frac{19}{20})}$$

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{\ln(0.95)}+\frac{\log_{0.95}\left(\frac{83}{450}\right)}{2}$$

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