Algebra - Solving Exponential Equations

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Section 6.3 : Solving Exponential Equations

2. Solve the following equation.

\[{5^{1 - x}} = 25\]

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Start Solution

Recall 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 2

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