Warm Up Solutions

Solve. Round to the nearest thousandth.
$12^{2 x - 3} = 15$

$\begin{array}{rcl} \ln 12^{2 x - 3} & = & \ln 15 \\ (2 x - 3) \ln 12 & = & \ln 15 \\ 2 x - 3 & = & \frac{\ln 15}{\ln 12} \\ 2 x & = & \frac{\ln 15}{\ln 12} + 3 \\ x & = & \frac{\ln 15}{2 \ln 12} + \frac{3}{2} \\ x & \approx & 2.045 \end{array}$

Note

Q: If you solved problem 1 using common logs, would you get the same answer?

A: Yes, the solutions would be equivalent.

Q: In which step is it best to round in problem 1?

A: Always round as the last step in the problem to ensure the greatest accuracy.

Determine if f(x) and g(x) are inverses. 
$f (x) = {(\frac{1}{2})}^{x + 5} g (x) = - \log_{2} x - 5$

$\begin{array}{rcl} (g \circ f) (x) & = & - \log_{2} {(\frac{1}{2})}^{x + 5} - 5 \\ = & - \log_{2} 2^{- (x + 5)} - 5 \\ = & - (- (x + 5)) - 5 \\ = & x + 5 - 5 \\ = & x \end{array}$

$\begin{array}{rcl} (f \circ g) (x) & = & {(\frac{1}{2})}^{- \log_{2} x + 5 - 5} \\ = & {(\frac{1}{2})}^{- {\log_{2}}^{x}} \\ = & {(2)}^{- 1 (- \log_{2} x)} \\ = & {(2)}^{\log_{2} x} \\ = & x \end{array}$

Yes, f(x) and g(x) are inverses of one another.