Last Updated Sep 27, 2025

Simplifying Expressions Simplified Revision Notes for Leaving Cert Mathematics

Revision notes with simplified explanations to understand Simplifying Expressions quickly and effectively.

Learn about Logarithms for your Leaving Cert Mathematics Exam. This Revision Note includes a summary of Logarithms for easy recall in your Mathematics exam

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Simplifying Expressions

Recall the rules of logs :

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$$\log_a(xy)=\log_ax+\log_ay$$ $$\log_a\left(x^q\right)=q\log_ax$$ $$\log_a\left(\tfrac{1}{x}\right)=-\log_ax$$ $$a^{\log_ax}=x$$ $$\log_a\left(\tfrac{x}{y}\right)=\log_ax-\log_ay$$ $$\log_a1=0$$ $$\log_a\left(a^x\right)=x$$

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Example

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Write as a single log and simplify :

$$\log_x2^x+\log_x2^{-x}$$

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$$\begin{align*} \log_x2^x+\log_x2^{-x} &= x\log_x2+-x\log_x2^{} & \text{\footnotesize\textcolor{gray}{\( \log_a\left(x^q\right)=q\log_ax \)}} \\\\ &= :highlight[0] & \end{align*}$$

Example

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If $\log_{10}x=(1+c)$ and $\log_{10}y=(1-c)$, show that $xy=:highlight[100].$

Our end goal is to derive $xy=:highlight[100]$, which contains no $c$ term, which implies we need to eliminate the $c$ term :

$$\begin{align*} \log_{10}(x)&=(1+c) & \\ \log_{10}(x)-1 &= c & \text{\footnotesize\textcolor{gray}{(\( -1 \))}} \\\\ \log_{10}(y)&=(1-c) & \\ \log_{10}(y)-1&=-c & \text{\footnotesize\textcolor{gray}{(\( -1 \))}} \\ -\log_{10}(y)+1&=c & \text{\footnotesize\textcolor{gray}{(\( \cdot1 \))}} \end{align*}$$

We have two separate expressions with $c$ isolated, so we can equate them

$$\begin{align*} \log_{10}(x)-1 &= -\log_{10}(y)+1 & \\ \log_{10}(x)-1+\log_{10}(y)-1 &= 0 \\ \log_{10}(x)+\log_{10}(y)-2 &= 0 & \\ \log_{10}(xy)-2 &= 0 & \\ \log_{10}(xy) &= 2 \\ xy &= 10^2 =:highlight[100]& \end{align*}$$

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