Simplifying Expressions (Leaving Cert Mathematics): Revision Notes

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Simplifying Expressions

Recall the rules of logs :

log⁡a(xy)=log⁡ax+log⁡ay\log_a(xy)=\log_ax+\log_ay log⁡a(xq)=qlog⁡ax\log_a\left(x^q\right)=q\log_ax log⁡a(1x)=−log⁡ax\log_a\left(\tfrac{1}{x}\right)=-\log_ax alog⁡ax=xa^{\log_ax}=x log⁡a(xy)=log⁡ax−log⁡ay\log_a\left(\tfrac{x}{y}\right)=\log_ax-\log_ay log⁡a1=0\log_a1=0 log⁡a(ax)=x\log_a\left(a^x\right)=x

Example

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Write as a single log and simplify :

log⁡x2x+log⁡x2−x\log_x2^x+\log_x2^{-x}
log⁡x2x+log⁡x2−x=xlog⁡x2+−xlog⁡x2log⁡a(xq)=qlog⁡ax=:highlight[0] \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⁡10x=(1+c)\log_{10}x=(1+c) and log⁡10y=(1−c)\log_{10}y=(1-c), show that xy=:highlight[100].xy=:highlight[100].

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

log⁡10(x)=(1+c)log⁡10(x)−1=c(−1)log⁡10(y)=(1−c)log⁡10(y)−1=−c(−1)−log⁡10(y)+1=c(⋅1) \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

log⁡10(x)−1=−log⁡10(y)+1log⁡10(x)−1+log⁡10(y)−1=0log⁡10(x)+log⁡10(y)−2=0log⁡10(xy)−2=0log⁡10(xy)=2xy=102=:highlight[100] \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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