Exponential Equations using Logarithms
We can find solutions to simple exponential equations where we can make equal bases and then equate exponents (indices).
For example, $2^{x}=8$ can be written as $2^x = 2^3$. Therefore the solution is $x=3$.
However, it is not always easy to make the same bases, such as $2^x=5$. We use $\textit{logarithms}$ to find the exact solution.
If the base is not the same and the numbers cannot be written with the same base then logarithms can be used.
It is possible to take the logarithm of both sides of an equation, provided the same base is used.
Example 1
Solve $10^x = 9$ for $x$, correct to $3$ decimal places.
\( \begin{align} \displaystyle
10^x &= 9 \\
x &= \log_{10}{9} \\
x &= 0.95424 \cdots &\text{using a calculator}\\
\therefore x &= 0.954
\end{align} \)
Example 2
Solve $2^{x+1} = 12$ for $x$, correct to $3$ decimal places.
\( \begin{align} \displaystyle
2^{x+1} &= 12 \\
\log_{10}{2^{x+1}} &= \log_{10}{12} \\
(x+1)\log_{10}{2} &= \log_{10}{12} \\
x+1 &= \dfrac{\log_{10}{12}}{\log_{10}{2}} \\
x &= \dfrac{\log_{10}{12}}{\log_{10}{2}} -1 \\
x &= 2.58496 \cdots \\
\therefore x &= 2.585
\end{align} \)
Example 3
Solve $\Big(\dfrac{1}{2}\Big)^x = 7$ for $x$, correct to $3$ decimal places.
\( \begin{align} \displaystyle
\Big(\dfrac{1}{2}\Big)^x &= 7 \\
(2^{-1})^x &= 7 \\
2^{-x} &= 7 \\
\log_{10}{2^{-x}} &= \log_{10}{7} \\
-x\log_{10}{2} &= \log_{10}{7} \\
-x &= \dfrac{\log_{10}{7}}{\log_{10}{2}} \\
x &= -\dfrac{\log_{10}{7}}{\log_{10}{2}} \\
x &= -2.80735 \cdots &\text{using a calculator}\\
\therefore x &= -2.807
\end{align} \)
Example 4
Solve \( 12 \times 5^{0.08x} = 72 \) for \( x \), correct to \( 3 \) significant figures.
\( \begin{align} \displaystyle
5^{0.08x} &= 6 \\
\log_{10}{5^{0.08x}} &= \log_{10}{6} \\
0.08x\log_{10}{5} &= \log_{10}{6} \\
0.08x &= \dfrac{\log_{10}{6}}{\log_{10}{5}} \\
x &= \dfrac{1}{0.08} \times \dfrac{\log_{10}{6}}{\log_{10}{5}} \\
x &= 13.91603 \cdots \\
\therefore x &= 13.9
\end{align} \)
Example 5
Rearrange $y=25 \times 2^{0.5x}$ to give $x$ in terms of $y$.
\( \begin{align} \displaystyle
25 \times 2^{0.5x} &= y \\
2^{0.5x} &= \dfrac{y}{25} \\
0.5x &= \log_{2}{\dfrac{y}{25}} \\
x &= \dfrac{1}{0.5}\log_{2}{\dfrac{y}{25}} \\
\therefore x &= 2\log_{2}{\dfrac{y}{25}}
\end{align} \)
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