Algebra - Solving Exponential Equations

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Section 6.3 : Solving Exponential Equations

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2. Solve the following equation.

${5^{1 - x}} = 25$

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