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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 : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 1}}\).