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