Logarithms are the inverse of exponentials. This can be useful in solving for the exponent of an equation in the form $x=b^y$ or vice versa since the exponential function $x=b^y$ can be written in the form of a logarithmic function $y=\log_{b}x$.
$$\begin{align} y=\log_{b}x \quad\text{is equivalent to}\quad x=b^y \end{align}$$
Where $b$ is any number given that $b>0$ and $b \neq 1$. And $y=\log_{b}x$ is usually read as “log base $b$ of $x$”
If you recall the methods in solving for the inverse of a function, let us attempt to solve the inverse of the exponential function $f(x)=b^x$,
\[\begin{align*} f(x) &= b^x \\ y &= b^x && \tcA{\text{replace $f(x)$ with $y$}} \\ x &= b^y && \tcA{\text{switch $x$ and $y$}} \\ y &= \log_b{x} && \tcA{\text{solve for $y$}} \\ f^{-1}(x) &= \log_b{x} \end{align*}\]Hence, this proves that Exponential and Logarithmic Functions are indeed inverses of each other.
$\example{1}$ Solve for $x$ if $\log_{10} x =2$
$\solution$
Express the logarithmic function as an exponential function,
\[\begin{align*} \log_{10} x &= 2 \\ x &= 10^2 \quad \tcA{\log_b x = y \;\tcB{\Rightarrow}\; x=b^y} \\ &= 100 \tagans \end{align*}\]Common Logarithm (Briggsian Logarithm)
Common Logarithm is a type logarithm whose base is 10 and is the inverse of the exponential function $10^x$. It is abbreviated as “log” which means the same as “$\log_{10}$”. The function $x= 10^y$ can be expressed in the logarithmic form $y=\log_{10}x$ which can also be written as $y=\log{x}$.
$$\begin{align} \log{x} = \log_{10}x \end{align}$$
Natural Logarithm (Napierian Logarithm)
Natural logarithm is a type logarithm whose base is $e$ (Euler’s number) and is the inverse of the exponential function $e^x$. It is abbreviated as “ln” which means the same as “$\log_{e}$”. The function $x=e^y$ can be expressed in the logarithmic form $y=\log_{e}x$ which can also be written as $y=\ln{x}$.
$$\begin{align} \ln{x} = \log_{e}x \end{align}$$
Properties of Common and Natural Logarithm
$\tcBal{Properties of Common Logarithms}$ $$ \begin{align} \log(xy) &= \log x + \log y \\ \log\br{\frac{x}{y} } &= \log x - \log y \\ \log x^n &= n \log x \\ \log_yx &= \frac{\log x}{\log y} \\ \log_x x &= 1 \\ \log 1 &= 0 \\ y^{\log_{y}x} &= x \\ \log 0 &= -\infty \end{align} $$
$\tcBal{Properties of Natural Logarithms}$ $$ \begin{align} \ln(xy) &= \ln x + \ln y \\ \ln\br{ \frac{x}{y} } &= \ln x - \ln y \\ \ln x^n &= n \ln x \\ \log_yx &= \frac{\ln x}{\ln y} \\ \ln e &= 1 \\ \ln 1 &= 0 \\ e^{\ln x} &= x \\ \ln 0 &= -\infty \end{align} $$
$\example{2}$ Solve for $x$: $\log_4(x+3) = 2$
$\solution$
Express the logarithmic function in an exponential form,
\[\begin{align*} \log_4(x+3) &= 2 \\ x+3 &= 4^2 \quad\quad \tcA{\log_b x = y \;\tcB{\Rightarrow}\; x=b^y} \\ x &= 16-3 \\ &= 13 \tagans \end{align*}\]$\example{3}$ Solve for $x$: $\log_{5}(x+3)+\log_{5}5 = 2$
$\solution$
Combine the terms that contains logarithms,
\[\begin{align*} \log_{5}(x+3)+\log_{5}5 &= 2 \\ \log_{5}[5(x+3)] &= 2 \quad \tcA{\log(xy)=\log{x} + \log{y}} \\ 5(x+3) &= 5^2 \\ 5x + 15 &= 25 \\ x &= 2 \tagans \end{align*}\]$\example{4}$ Solve for $x$: $\ln x = \ln\frac{1}{2}+\ln 2-\ln 3$
$\solution$
Combine the terms for each side of the equation that contains logarithms.
\[\begin{align*} \ln x &= \ln\frac{1}{2}+\ln 2-\ln 3 \\ &= \ln\br{\frac{\frac12 \cdot 2}{3}} \qquad\tcA{\ln\br{\frac{x}{y}} = \ln x - \ln y} \\ \ln x &= \ln\frac{1}{3} \end{align*}\]Raise both sides to a power with base $e$,
\[\begin{align*} e^{\ln{x}} &= e^{\ln\frac13} \qquad\qquad\quad\tcA{e^{\ln x} = x} \\ x &= \frac{1}{3} \tagans \end{align*}\]$\example{5}$ Solve for $x$: $\ln{x} = 2\ln{3}-\ln{e}$
$\solution$
\[\begin{align*} \ln{x} &- 2\ln{3} - \ln{e} \\ &= \ln{3^2}-\ln{e} \quad\tcA{\ln{x^n} = n\ln x} \\ &= \ln\frac{3^2}{e} \\ e^{\ln x} &= e^{\ln\frac{9}{e}} \\ x &= \frac{9}{e} \tagans \end{align*}\]
$\example{6}$ Solve for $x$: $\ln(x+6)+\ln(2-x) = \ln 15$
$\solution$
\[\begin{align*} \ln(x+6)+\ln(2-x) &= \ln 15 \\ \ln[(x+6)(2-x)] &= \ln{15} \\ \ln(-x^2-4x+12) &= \ln{15} \\ e^{\ln(-x^2-4x+12)} &= e^{\ln{15}} \\ -x^2-4x+12 &= 15 \\ x^2+4x+3 &= 0 \end{align*}\]
Then factor the equation,
\[\begin{align*} (x+3)(x+1) &= 0 \end{align*}\]Thus,
\[\begin{align*} x+3 &= 0 & x+1 &= 0 \\ x &= -3 & x &= -1 \tagans \end{align*}\]$\example{7}$ Solve for $x$: $2^x = 3$
$\solution$
Express the exponential function in a logarithmic form,
\[\begin{align*} 2^x &= 3 \\ x &= \log_{2}3 \tagans \end{align*}\]$\example{8}$ Solve for $x$: $3^{5x} = 4^{2x+1}$
$\solution$
Get the natural logarithm of both sides of the equation,
\[\begin{align*} 3^{5x} &= 4^{2x+1} \\ \ln(3^{5x}) &= \ln(4^{2x+1}) \\ 5x \ln{3} &= (2x+1)\ln{4} \\ 5x \ln{3} &= 2x \ln{4} + \ln{4} \\ x( 5 \ln{3} - 2 \ln{4} ) &= \ln{4} \\ x &= \frac{\ln{4}}{5 \ln{3} - 2 \ln{4}} \tagans \end{align*}\]$\example{9}$ Solve for $x$: $e^x-6e^{-x} = -1$
$\solution$
To remove the negative exponent, multiply both sides of the equation by $e^x$,
\[\begin{align*} e^x-6e^{-x} &= -1 \\ e^x (e^x-6e^{-x}) &= e^x(-1) \\ e^{2x}-6 &= -e^x \\ e^{2x}+e^x-6 &= 0 \end{align*}\]The factor the equation,
\[\begin{align*} (e^x+3)(e^x-2) &= 0 \end{align*}\]Thus,
\[\begin{align*} x+3 &= 0 & x-2 &= 0 \\ x &= -3 & x &= 2 \tagans \end{align*}\]