Exponential Inequalities using Logarithms


Inequalities are worked in exactly the same way except that there is a change of sign when dividing or multiplying both sides of the inequality by a negative number.

\begin{array}{|c|c|c|} \hline
\log_{2}{3}=1.6>0 & \log_{5}{3}=0.7>0 & \log_{10}{3}=0.5>0 \\ \hline
\log_{2}{2}=1>0 & \log_{5}{2}=0.4>0 & \log_{10}{2}=0.3>0 \\ \hline
\log_{2}{1}=0 & \log_{5}{1}=0 & \log_{10}{1}=0 \\ \hline
\log_{2}{0.5}=-1<0 & \log_{5}{0.5}=-0.4<0 & \log_{10}{0.5}=-0.3<0 \\ \hline \log_{2}{0.1}=-3.3<0 & \log_{5}{0.1}=-1.4<0 & \log_{10}{0.1}=-1<0 \\ \hline \end{array} Thus we can find the following properties. \begin{array}{|c|c|c|} \hline a<1 & a=1 & a>1 \\ \hline
\log_{x}{a} <0 & \log_{x}{a}=0 & \log_{x}{a} >0 \\ \hline
\end{array}

$\textit{Case 1}$

\( \begin{align} \displaystyle
x\log_{10}{2} &> 5 \\
x &\color{green}> \dfrac{5}{\log_{10}{2}} \\
\end{align} \)

$\textit{Case 2}$

\( \begin{align} \displaystyle
x\log_{10}{0.5} &> 5 \\
x &\color{red}< \dfrac{5}{\log_{10}{0.5}} \\ \end{align} \)

Example 1

Solve $2^x > 9$ for $x$, correct to $3$ significant figures.

\( \begin{align} \displaystyle
2^x &> 9 \\
\log_{10}{2^x} &> \log_{10}{9} \\
x\log_{10}{2} &> \log_{10}{9} \\
x &> \dfrac{\log_{10}{9}}{\log_{10}{2}} \\
x &> 3.1699 \cdots \\
\therefore x &> 3.17 \\
\end{align} \)

Example 2

Solve $0.5^x \le 3$ for $x$, correct to $3$ significant figures.

\( \begin{align} \displaystyle
0.5^x &\le 3 \\
\log_{10}{0.5}^x &\le \log_{10}{3} \\
x\log_{10}{0.5} &\le \log_{10}{3} \\
x &\ge \dfrac{\log_{10}{3}}{\log_{10}{0.5}} \\
x &\ge -1.5849 \cdots \\
\therefore x &\ge -1.58 \\
\end{align} \)


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