Solving Equations with Indices

In some cases you may get equations with unknown powers on both sides, such as this one :

2^{x+8}=2^{5x}

Since the bases are equal, you can cancel them out, leaving only the powers to work with :

\cancel{2}^{x+8}=\cancel{2}^{5x}

\begin{align*} {x+8}&=5x \\ {8}&=4x \\ 2&=x \end{align*}

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Why does this work ? You can take the logarithm of both sides to eliminate the base.

\begin{align*} a^x&=a^y \\ \log_a(a^x)&=\log_a(a^y) \\ x&=y \end{align*}

Example

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Solve the equation :

2^{x^2}=8^{2x+9}

\begin{align*} 2^{x^2}&=\left(2^{3}\right)^{2x+9} &\text{\footnotesize\textcolor{gray}{(\(\text{Rewrite as } 2^a \))}} \\\\ 2^{x^2}&=2^{6x+27} &\text{\footnotesize\textcolor{gray}{\(\left( a^p\right)^q=a^{pq}\)}} \\\\ {x^2}&={6x+27} &\text{\footnotesize\textcolor{gray}{(\(\text{Cancel bases}\))}} \\\\ 0&={-x^2+6x+27} & \\\\ 0&=(x+3)(x-9) \\\\ x&=-3,x=9 \end{align*}

In some cases, rewriting the base isn't so simple. A powerful technique we can use would be to encapsulate the whole power term into something simpler i.e. using substitution.

Example

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Solve the equation :

2^{2y+1}-5(2^y)+2=0

Let $2^y=a$ .

\begin{align*} 2^{2y}\cdot2^1-5(2^y)+2&=0 &\text{\footnotesize\textcolor{gray}{\( a^{p+q}=a^p\cdot a^q \)}} \\\\ \left(2^y \right)^2\cdot2^1-5(2^y)+2&=0 &\text{\footnotesize\textcolor{gray}{\(\left( a^p\right)^q=a^{pq}\)}} \\\\ \left(a \right)^2\cdot2^1-5(a)+2&=0 &\text{\footnotesize\textcolor{gray}{\(2^y=a\)}} \\\\ 2a^2-5a+2&=0 & \\\\ a=2,a&=\tfrac{1}{2} & \\ \end{align*}

We've solved for $a$ , but we just used $a$ to make the solving easier for us, now resubstitute the original expression, recall $2^y=a$ .

\begin{align*} 2^y&=2 \\ y&=1 \end{align*}

\begin{align*} 2^y&=\tfrac{1}{2} \\ y &=-1 \end{align*}

Solving Equations with Indices Simplified Revision Notes for Leaving Cert Mathematics

Solving Equations with Indices

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