Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 6.3 : Solving Exponential Equations
2. Solve the following equation.
\[{5^{1 - x}} = 25\]Show All Steps Hide All Steps
Start SolutionRecall the property that says if \({b^x} = {b^y}\) then \(x = y\). In this case it looks like we can’t use this property. However, recall that \(25 = {5^2}\) and if we write the right side of our equation using this we get,
\[{5^{1 - x}} = {5^2}\]Now each exponential has the same base, 5 to be exact, so we can use this property to just set the exponents equal. Doing this gives,
\[1 - x = 2\] Show Step 2Now all we need to do is solve the equation from Step 1 and that is a simple linear equation. Here is the solution work.
\[\begin{align*}1 - x & = 2\\ - x & = 1\hspace{0.25in} \to \hspace{0.25in}x = - 1\end{align*}\]So, the solution to the equation is then : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 1}}\).