### Exponential functions and logarithms: Logarithms

### Calculation rules for logarithmic functions

Calculation rules The calculation rules for the natural logarithm, which we have already seen before, apply as well for logarithmic functions with any base. They can be derived from the properties of exponential functions. The following equalities are valid for all positive real numbers \(x\) and \(y\), and for all bases \(g\) and \(h\), and for every rational number \(r\): \[\begin{aligned}

\log_g(x\cdot y) &= \log_g(x)+\log_g(y)\\ \\

\log_g\!\left(\frac{x}{y}\right) &=\log_g(x)-\log_g(y) \\ \\

\log_g(x^r) &= r\cdot \log_g(x) \\ \\

\log_g(x) &=\frac{\log_h(x)}{\log_h(g)}

\end{aligned}\]

Via the last formula you can convert logarithms with base \(g\) convert into logarithms with another base. For example: \[\log_{10}(x) =\frac{\log_e(x)}{\log_e(10)}=\frac{\ln(x)}{\ln(10)}\]

To illustrate the calculation rules, we give an example of consistent application of the rules to bring some formula manipulation to a successful end.

express \(y\) as a function of \(x\) and simplify such that no logarithm is present anymore.

We first write the right-hand hand side of the given formula as a base-10 logarithm: \[\begin{aligned} -3+\frac{1}{2}\cdot\log(x) &= \log(10^{-3})+ \log(x^{\frac{1}{2}})\\ &=\log(10^{-3}\cdot x^{\frac{1}{2}})\end{aligned}\] So \[\log(y)= \log(10^{-3}\cdot x^{\frac{1}{2}})\] It follows: \[\begin{aligned}y&=10^{-3}\cdot x^{\frac{1}{2}} \\&=0.001\cdot \sqrt{x}\end{aligned}\]