Logarithms are a type of functions that are very useful.

**Definition**

A logarithm is the inverse of an exponent. So:

\[ x^a = y \iff \log_xy=a \]

\(\log_xy\) is read as: "Log \(y\) base x", the log function basically asks, \(x\) raised to which power will give \(y\)?

Note that \(x\) is greater than 0 and cannot be 1 and \(y\) needs to be positive

**Logarithmic Identities**

**Identity 1**: \(\log_a 1 = 0\)

The First Identity comes from a property of exponents. Any number raised to the power of 0 is 1 (except for \(0^0\), that is undefined) so \(\log_a 1 = 0 \)

**Identity 2**: \(\log_a a = 1\)

For all \(a\) \[ a^1=a \] Using the Definition of Logarithms: \[ \therefore \log_a a = 1 \blacksquare\]

**Identity 3**: \(\log_a xy = \log_a x +\log_ay \)

Let \( \log_a x = u\) and \( \log_ay=v\)

Then \(a^u =x\) and \(a^v = y)

\[xy = a^ua^v = a^{u+v}\]

Taking the Log of both sides:

\[ \begin{array} {l l } \log_a xy & = \log_a a^{u+v} \\ & = u+v\\ & = \log_ax+\log_ay \blacksquare \end{array} \]

**Identity 4** : \(\log_a \frac xy = \log_a x - \log_ay \)

Let \( \log_a x = u\) and \( \log_ay=v\)

Then \(a^u =x\) and \(a^v = y)

\[\frac xy = \frac{a^u}{a^v} = a^{u-v}\]

Taking the Log of both sides:

\[ \begin{array} { l l } \log_a \frac xy & = \log_a a^{u-v} \\ & = u-v \\ &= \log_ax-\log_ay \blacksquare \end{array} \]

**Identity 5:** \(\log_a x^n = n\log_ax\)

Let \(\log_ax=u\)

Then according to the definition of Logarithms: \(a^u=x\)

By Raising both sides to the power of n: \[x^n = (a^u)^n = a^{nu}\]

Take the log of both sides: \[ \begin{array} {l l } \log_a x^n & = \log_a a^{nu}\\ &= nu\\ &= n\log_ax \blacksquare \end{array} \]

This Identity is very important, it is very helpful in problems where something is raised to the power of a variable, you can just take the log and then "take" the variable down and the log term becomes a constant

**Identity 6**: \(\log_ax=\frac{\log_bx}{\log_ba}\)

Let \(log_ax=u\)

Then \(a^u=x\)

Taking the log of both sides with a base of a completely new number \(b\):

\[ \log_ba^u=\log_bx \]

Using Identity 5 which we proved earlier:

\[ \begin{array} {l l} u\log_ba & =\log_bx \\ u& =\frac{\log_bx}{\log_ba} \\ \log_ax & =\frac{\log_bx}{\log_ba}\blacksquare \end{array} \]

**Special Logarithms**

If a logarithm doesn't have a base (like \(\log\)) then it is called the "Common Logarithm", it is a logarithm in base 10

\(e\) is a special number that is approximately 2.71828... \(\log_e x\) is often written as \(\ln x\) This is called the Natural Logarithm

**Why do we need logarithms?**

When Scientists do experiments their results often do not have a linear relationship but instead have exponential relationships that often aren't recognisable easily, but when the logarithms of the results are taken the results will often show linear relationships.

Example:

A Radioactive source has a half life of 1 year, how many years will it take for the radioactive source to have only one tenth of its radioactive elements remaining?

Let \(N_0\) be the number of radioactive elements in the beginning.

\[\frac 1{10} N_0 = N_0\left(\frac 12\right)^n\] \[ \frac 1{10} = \left(\frac 12\right)^n\]

Before you learnt logarithms, you wont know how to solve this, as the \(n\) is at the top, but since we know logs, we can solve it like this:

\[\log \frac 1{10} = \log\left(\frac 12\right)^n\] \[-1 = n\log\left(\frac 12\right)\] \[n = \frac{-1}{\log\left(\frac 12\right)}\]

Using a calculator: \[n = \frac{-1}{-0.301}\] \[n =3.22\]

It will take approximate 3.22 years

## Comments

Sort by:

TopNewestHere's another example, taken from my local math team:

– Michael Tang · 3 years, 5 months agoLog in to reply

\[\log_a (b^5) = 3\]

\[5\log_a b =3\]

\[\log_a b =\frac{3}{5}\]

Applying \(\log\) on RHS,

\[\log _a b = \log_a a^{\frac{3}{5}}\]

Canceling \(\log\) from both sides

\[b = a^{\frac{3}{5}}\]

\[b^{\frac{5}{3}} = a\]

Therefore,

\[\log_b a^5 = \log_b (b^{\frac{5}{3}})^5\]

\[\log_b a^5 = \log_b b^{\frac{25}{3}}\]

\[\log_b a^5 = \frac{25}{3}\log_b b\]

\[\log_b a^5 = \frac{25}{3}\blacksquare\] – 敬全 钟 · 3 years, 5 months ago

Log in to reply

– Sadman Sakib · 3 years, 5 months ago

How about using \( \displaystyle \log_b a = \frac {1}{\log_a b}\) ?Log in to reply

\[\frac{1}{\log_a b^5} = \log_{b^5}{a} = \frac{1}{5}\log_b a = \frac{1}{3}\].

Then,

\[\log_b a =\frac{5}{3}\]

\[\implies \log_b a^5 = 5\log_b a = \frac{25}{3}\blacksquare\] – 敬全 钟 · 3 years, 5 months ago

Log in to reply

– Phanindra Sarma · 3 years, 5 months ago

loga(b^5)=3. Implies a^3=b^5. Implies, a=b^5/3. Therefore a^5=b^25/3. From this we get 25/3.Log in to reply

can anyone tell me how to solve a log within a log..log a ^(log b^a)/log b^(log a^b).. – Writo Mukherjee · 2 years, 12 months ago

Log in to reply

Actually, the base of the logarithm, \(x,\) does not have to be greater than \(1.\) It may also be from \(0\) to \(1.\) However, these logarithms are admittedly not as nice, and are hardly necessary: for example, \(\log_{1/2} y\) is the same thing as \(-\log _2 y.\) – Alexander Borisov · 3 years, 5 months ago

Log in to reply

what is the relation that is used here – Anil Kumar · 3 years, 5 months ago

Log in to reply

logarithms, i had seen them in physics from last year, taking log then antilogs, it was confusing the way our school teachers taught us, so i used the concept without using "log" word.

eg. 23.86 x 3.14 = 10^(1.377) x 10^(0.497) = 10^(1.377 + 0.497) = 10^1.874 = 74.81

Log in to reply

– Vinay Pandey · 3 years, 5 months ago

Think logarithm in terms of function and expansion, you will get clear idea.Log in to reply

I didn't understand the right side of the equation in the radioactive element example. What is (1/2)^n ? Please explain – Muhammad Khan · 3 years, 5 months ago

Log in to reply

– Yan Yau Cheng · 3 years, 5 months ago

Every year the number of elements will decrease by half, n is the number of yearsLog in to reply

– Muhammad Khan · 3 years, 5 months ago

Thanks! :DLog in to reply