Algebra - Solving Exponential Equations

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Section 6.3 : Solving Exponential Equations

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3. Solve the following equation.

\[{8^{{x^{\,2}}}} = {8^{3x + 10}}\]

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Start Solution

Recall the property that says if \({b^x} = {b^y}\) then \(x = y\). Since each exponential has the same base, 8 in this case, we can use this property to just set the exponents equal. Doing this gives,

\[{x^2} = 3x + 10\] Show Step 2

Now all we need to do is solve the equation from Step 1 and that is a quadratic equation that we should be able to quickly solve. Here is the solution work.

\[\begin{align*}{x^2} & = 3x + 10\\ {x^2} - 3x - 10 & = 0\\ \left( {x - 5} \right)\left( {x + 2} \right) & = 0\hspace{0.25in} \to \hspace{0.25in}x = - 2,\,\,\,\,\,x = 5\end{align*}\]

So, the solutions to the equation are then : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 2}}\) and \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 5}}\).