## Exponential and Logarithmic Equations

Many of the same principles of equation solving extend to equations that contain either: 1) exponential expressions where the variable appears in the exponent, or 2) logarithmic expressions.

In particular, one should remember that exponential functions of a given base (e.g., $f(x) = b^x$) and logarithmic functions of the same base (e.g., $f(x) = \log_b x$) are inverses of one another. Each can be used to eliminate the other as the next two examples show:

**Example 1**

$$\begin{array}{rclc}
20 \left( \displaystyle{\frac{1}{2}} \right)^{x/3} &=& 5 & \scriptsize{\textrm{There is only one occurrence of } x \textrm{ so, we isolate it by applying inverse operations...}}\\\\
\left( \displaystyle{\frac{1}{2}} \right)^{x/3} &=& \displaystyle{\frac{1}{4}} & \scriptsize{\textrm{first, dividing by 20 to undo multiplication by 20}}\\\\
\log_{1/2} \left( \displaystyle{\frac{1}{2}} \right)^{x/3} &=& \log_{1/2} \displaystyle{\frac{1}{4}} & \scriptsize{\textrm{then, taking a log base } 1/2 \textrm{ to undo exponentiation base} 1/2}\\\\
\displaystyle{\frac{x}{3}} &=& 2 &\\\\
x &=& 6 & \scriptsize{\textrm{finally, multiplying by 3 to undo a division by 3}}
\end{array}$$

**Example 2**

$$\begin{array}{rclc}
3 \log_5 x^3 &=& 9 & \scriptsize{\textrm{There is only one occurrence of } x \textrm{ so, we isolate it by applying inverse operations...}}\\\\
\log_5 x^3 &=& 3 & \scriptsize{\textrm{first, by dividing by 3 to undo a multiplication by 3}}\\\\
5^{\log_5 x^3} &=& 5^3 & \scriptsize{\textrm{then, using both sides as an exponent on 5, to undo a log base 5}}\\\\
x^3 &=& 5^3 & \\\\
x &=& 5 & \scriptsize{\textrm{finally, taking a cube root of both sides to undo the cubing of } x}
\end{array}$$

Note, one can use the properties of logarithms to consolidate several logarithms into a single logarithm. This frequently makes solving the related equation easier.

**Example 3**

$$\begin{array}{rclc}
\log x + \log (x+3) &=& 1 & \scriptsize{\textrm{recall a sum of logs (with the same base) can be written as a log of a product...}}\\\\
\log x(x+3) &=& 1 & \scriptsize{\textrm{now, we eliminate the logarithm by using both sides as exponents on a base of } 10}\\\\
10^{\log x(x+3)} &=& 10^1 & \scriptsize{\textrm{then, using both sides as an exponent on 10, to undo a log base 10}}\\\\
x(x+3) &=& 10 & \scriptsize{\textrm{from here, it is a familiar quadratic -- so we proceed to solve in the normal way...}}\\\\
x^2 + 3x - 10 &=& 0 & \scriptsize{\textrm{finally, taking a cube root of both sides to undo the cubing of } x}\\\\
(x+5)(x-2) &=& 0 &\\\\
x &=& 2 \textrm{ or } -5 & \scriptsize{\textrm{properties of logs can produce extraneous solutions -- so we must check these!}}\\\\
x &=& 2 & \scriptsize{\textrm{noting that } -5 \textrm{ is a domain issue in the original equation}}
\end{array}$$

**Example 4**

$$\begin{array}{rclc}
\ln (4x+6) - \ln (x+5) &=& \ln x & \scriptsize{\textrm{recall a difference of logs (with the same base) can be written as a log of a quotient...}}\\\\
\ln \displaystyle{\frac{4x+6}{x+5}} &=& \ln x & \scriptsize{\textrm{now, we eliminate the logs by using both sides as exponents on a base of } e}\\\\
e^{\ln (4x+6)/(x+5)} &=& e^{\ln x} & \scriptsize{\textrm{then, using both sides as an exponent on 5, to undo a log base 5}}\\\\
\displaystyle{\frac{4x+6}{x+5}} &=& x & \scriptsize{\textrm{from here, it is just an equation involving rational expressions...}}\\\\
4x+6 &=& x(x+5) & \scriptsize{\textrm{eliminating the denominator, we reveal a quadratic equation}}\\\\
x^2+x-6 &=& 0 & \scriptsize{\textrm{which can be solved by factoring...}}\\\\
(x+3)(x-2) &=& 0 & \\\\
x &=& -3 \textrm{ or } 2 & \scriptsize{\textrm{properties of logs can produce extraneous solutions -- so we must check these!}}\\\\
x &=& 2 & \scriptsize{\textrm{noting that } -3 \textrm{ is a domain issue in the original equation}}
\end{array}$$

In some situations, we must slightly delay attempts to remove the exponential expressions by taking logarithms...

**Example 5**

$$\begin{array}{rclc}
e^x + e^{-x} - 6 &=& 0 & \scriptsize{\textrm{here, we first eliminate the negative exponent by multiplying by } e^x}\\\\
e^{2x} + 1 - 6e^x &=& 0 & \scriptsize{\textrm{this is quadratic in terms of } e^x \ldots}\\\\
(e^x)^2 - 6(e^x) + 1 &=& 0 & \scriptsize{\textrm{seeing that this does not factor nicely, we use the quadratic formula}}\\\\
e^x &=& \displaystyle{\frac{6 \pm 4\sqrt{2}}{2}} & \scriptsize{\textrm{then noting the common factor of } 2}\\\\
e^x &=& 3 \pm 2\sqrt{2} & \scriptsize{\textrm{finally, we turn our attention to eliminating the exponential by taking a natural log}}\\\\
\ln e^x &=& \ln (3 \pm 2\sqrt{2}) & \\\\
x &=& \ln (3 \pm 2\sqrt{2})
\end{array}$$

Logarithms can also be used as a tool to get a variable expression out of an exponent -- even when the base of the exponentiation and the base of the logarithm disagree -- as the next example shows:

**Example 6**

$$\begin{array}{rclc}
4^{x+3} &=& 3^{-x} & \scriptsize{\textrm{take a log (base 10) of both sides}}\\\\
\log 4^{x+3} &=& \log 3^{-x} & \scriptsize{\textrm{now we can appeal to one of the properties of logs to relocate the variable exponents}}\\\\
(x+3)\log 4 &=& -x \log 3 & \scriptsize{\textrm{recalling that } \log 4 \textrm{ and } \log 3 \textrm{are constants, this is a linear equation!}}\\\\
(\log 4 + \log 3) x&=& -3 \log 4 & \scriptsize{\textrm{which we solve in the normal way by isolating } x}\\\\
x &=& \displaystyle{\frac{-3\log 4}{\log 4 + \log 3}} &
\end{array}$$