Precalculus Examples

$Q = 1.5 e^{\frac{\ln (0.5)}{2480} \cdot (1000)}$

Step 1

Divide both sides by $1.5 e^{\frac{\ln (0.5)}{2480} \cdot (1000)}$ to remove it from the front of the logarithm.

$\frac{Q}{1.5 e^{\frac{\ln (0.5)}{2480} \cdot (1000)}} = \frac{1.5 e^{\frac{\ln (0.5)}{2480} \cdot (1000)}}{1.5 e^{\frac{\ln (0.5)}{2480} \cdot (1000)}}$

Step 2

Simplify each side of the equation.

$0.88164105 Q =$

Step 3

For logarithmic equations, $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ such that $x > 0$ , $b > 0$ , and $b \neq 1$ . In this case, $b = 10$ , $x =$ , and $y = 0.88164105 Q$ .

$b = 10$

$x =$

$y = 0.88164105 Q$

Step 4

Substitute the values of $b$ , $x$ , and $y$ into the equation $b^{y} = x$ .

$10^{0.88164105 Q} =$