Logarithms - An Introduction

Logarithms are a type of functions that are very useful.

Definition

A logarithm is the inverse of an exponent. So:

xa=y    logxy=a x^a = y \iff \log_xy=a

logxy\log_xy is read as: "Log yy base x", the log function basically asks, xx raised to which power will give yy?

Note that xx is greater than 0 and cannot be 1 and yy needs to be positive

Logarithmic Identities

Identity 1: loga1=0\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 000^0, that is undefined) so loga1=0\log_a 1 = 0

Identity 2: logaa=1\log_a a = 1

For all aa a1=a a^1=a Using the Definition of Logarithms: logaa=1 \therefore \log_a a = 1 \blacksquare

Identity 3: logaxy=logax+logay\log_a xy = \log_a x +\log_ay

Let logax=u \log_a x = u and logay=v \log_ay=v

Then au=xa^u =x and av=y)\[xy=auav=au+va^v = y) \[xy = a^ua^v = a^{u+v}

Taking the Log of both sides:

logaxy=logaau+v=u+v=logax+logay \begin{array} {l l } \log_a xy & = \log_a a^{u+v} \\ & = u+v\\ & = \log_ax+\log_ay \blacksquare \end{array}

**Identity 4** : logaxy=logaxlogay\log_a \frac xy = \log_a x - \log_ay

Let logax=u \log_a x = u and logay=v \log_ay=v

Then au=xa^u =x and av=y)\[xy=auav=auva^v = y) \[\frac xy = \frac{a^u}{a^v} = a^{u-v}

Taking the Log of both sides:

logaxy=logaauv=uv=logaxlogay \begin{array} { l l } \log_a \frac xy & = \log_a a^{u-v} \\ & = u-v \\ &= \log_ax-\log_ay \blacksquare \end{array}

**Identity 5:** logaxn=nlogax\log_a x^n = n\log_ax

Let logax=u\log_ax=u

Then according to the definition of Logarithms: au=xa^u=x

By Raising both sides to the power of n: xn=(au)n=anux^n = (a^u)^n = a^{nu}

Take the log of both sides: logaxn=logaanu=nu=nlogax \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: logax=logbxlogba\log_ax=\frac{\log_bx}{\log_ba}

Let logax=ulog_ax=u

Then au=xa^u=x

Taking the log of both sides with a base of a completely new number bb:

logbau=logbx \log_ba^u=\log_bx

Using Identity 5 which we proved earlier:

ulogba=logbxu=logbxlogbalogax=logbxlogba \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\log) then it is called the "Common Logarithm", it is a logarithm in base 10

ee is a special number that is approximately 2.71828... logex\log_e x is often written as lnx\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 N0N_0 be the number of radioactive elements in the beginning.

110N0=N0(12)n\frac 1{10} N_0 = N_0\left(\frac 12\right)^n 110=(12)n \frac 1{10} = \left(\frac 12\right)^n

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

log110=log(12)n\log \frac 1{10} = \log\left(\frac 12\right)^n 1=nlog(12)-1 = n\log\left(\frac 12\right) n=1log(12)n = \frac{-1}{\log\left(\frac 12\right)}

Using a calculator: n=10.301n = \frac{-1}{-0.301} n=3.22n =3.22

It will take approximate 3.22 years

Note by Yan Yau Cheng
5 years, 10 months ago

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Comments

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can anyone tell me how to solve a log within a log..log a ^(log b^a)/log b^(log a^b)..

Writo Mukherjee - 5 years, 5 months ago

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Actually, the base of the logarithm, x,x, does not have to be greater than 1.1. It may also be from 00 to 1.1. However, these logarithms are admittedly not as nice, and are hardly necessary: for example, log1/2y\log_{1/2} y is the same thing as log2y.-\log _2 y.

Alexander Borisov - 5 years, 10 months ago

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what is the relation that is used here

anil kumar - 5 years, 10 months ago

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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

  • above is possible if you got a log table :) but still we can sometimes aprox it.

Soham Zemse - 5 years, 10 months ago

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

Vinay Pandey - 5 years, 10 months ago

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I didn't understand the right side of the equation in the radioactive element example. What is (1/2)^n ? Please explain

Muhammad Khan - 5 years, 10 months ago

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Every year the number of elements will decrease by half, n is the number of years

Yan Yau Cheng - 5 years, 10 months ago

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Thanks! :D

Muhammad Khan - 5 years, 10 months ago

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

Let a,ba, b be positive real numbers such that loga(b5)=3.\log_a (b^5) = 3. Determine the value of logb(a5).\log_b (a^5).

Michael Tang - 5 years, 10 months ago

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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.

Phanindra Sarma - 5 years, 10 months ago

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First of all, we have

loga(b5)=3\log_a (b^5) = 3

5logab=35\log_a b =3

logab=35\log_a b =\frac{3}{5}

Applying log\log on RHS,

logab=logaa35\log _a b = \log_a a^{\frac{3}{5}}

Canceling log\log from both sides

b=a35b = a^{\frac{3}{5}}

b53=ab^{\frac{5}{3}} = a

Therefore,

logba5=logb(b53)5\log_b a^5 = \log_b (b^{\frac{5}{3}})^5

logba5=logbb253\log_b a^5 = \log_b b^{\frac{25}{3}}

logba5=253logbb\log_b a^5 = \frac{25}{3}\log_b b

logba5=253\log_b a^5 = \frac{25}{3}\blacksquare

敬全 钟 - 5 years, 10 months ago

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How about using logba=1logab \displaystyle \log_b a = \frac {1}{\log_a b} ?

Sadman Sakib - 5 years, 10 months ago

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@Sadman Sakib That's a great idea. We have

1logab5=logb5a=15logba=13\frac{1}{\log_a b^5} = \log_{b^5}{a} = \frac{1}{5}\log_b a = \frac{1}{3}.

Then,

logba=53\log_b a =\frac{5}{3}

    logba5=5logba=253\implies \log_b a^5 = 5\log_b a = \frac{25}{3}\blacksquare

敬全 钟 - 5 years, 10 months ago

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