Question

Give a brief description about logarithm.

Answer 1

Logarithms are used to make the long and complicated calculations easy.
Eg: $3^5=243\to$ We have a relation between 3, 5 and 243
Now, same relation between 3,5 and 243 can be written as
$lo{g}_{3}243=5\Rightarrow 243=3^5$
Generalising $\stackrel{a^b=c}{\left(Exponentialform\right)}\Rightarrow \stackrel{\left(logarithmicform\right)}{lo{g}_{a}c=b}\Rightarrow \stackrel{c=a^b}{\left(logofctobaseaisb\right)}$
Eg: $2^{-3}=\frac{1}{8}=0.125\to lo{g}_{2}0.125=-3$
$||{|}^{ly}(81{)}^{\frac{1}{2}}=9\Rightarrow lo{g}_{81}9=\frac{1}{2}1)x^0=1\Rightarrow lo{g}_{x}1=0Eg:lo{g}_{5}1=0lo{g}_{10}=0x^1=x\Rightarrow lo{g}_{x}x=1Eg:lo{g}_{5}5=1lo{g}_{10}10=1$
Ques: $lo{g}_{10}1000=3,lo{g}_{10}\frac{1}{1000}=-3$
$lo{g}_2(x^2-4)=5x=\pm 6,lo{g}_{x}64=\frac{3}{2}x=16lo{g}_7(2x^2-1)=2x=\pm 5$
Laws of logarithms
1) Product law $\to lo{g}_a(m\times n)=lo{g}_{a}m+lo{g}_{a}nlo{g}_a(m\times n)=k\Rightarrow (m\times n)=a^k\stackrel{p}{mcal}\stackrel{q_q}{n=a}\therefore RHS=P+qlo{g}_a(m\times n)=lo{g}_a(a^p\times a^q)=lo{g}_a\left(a^{p+q}\right)=km\times n=m+m+\dots \dots ntimesx{a}^{p+q}=a^{k}k=p+q\therefore lo{g}_a\left(a^{p+q}\right)=(p+q)$
LHS

2) Quotient law $lo{g}_a\left(\frac{m}{n}\right)=lo{g}_{a}m-lo{g}_{a}n=p-q$
$\frac{lo{g}_{a}m}{lo{g}_{a}n}\ne mloga$

3) Power law $lo{g}_a(m{)}^n=nlo{g}_{a}m$
LHS
$lo{g}_a{m}^n=lo{g}_{a}m\times m\times m\dots \dots ntimes=lo{g}_{a}m+lo{g}_{a}n+logm\dots \dots ntimes=n\times lo{g}_{a}m\begin{array}{c}\\ lo{g}_{a}m=p\Rightarrow m=a^p\\ lo{g}_{a}n=q\Rightarrow n=a^q\end{array}|\begin{array}{c}\\ let,lo{g}_a\left(\frac{m}{n}\right)=K\\ \frac{m}{n}=a^k\\ \Rightarrow \frac{a^p}{a^q}=a^{p-q}=a^k\\ \therefore k=p-q\\ \therefore lo{g}_a\left(\frac{m}{n}\right)=p-q=RHS\end{array}$
Logarithms to the base 10 are called common logarithms
If no base is given - take it as 10
Expansion of expression using logarithms
$y=\frac{a^5\times b^3}{e^2}$
$logy=5loga+3logb-2logc$ (logarithmic expansion)
Ques: $\to 2lo{g}_{10}x+1=lo{g}_{10}250x=?$
$\Rightarrow lo{g}_{10}{x}^2=lo{g}_{10}25\therefore x^2=25x=\pm 5=5$
$lo{g}_{10}-5=x-5={10}^{x}lo{g}_{-10}-5=x-5=(-10{)}^2$

Solve for $x\to lo{g}_{x}49-lo{g}_{x}7+lo{g}_x\frac{1}{3\times 3}+2=0lo{g}_{x}49-lo{g}_{x}7-lo{g}_{x}343+lo{g}_x{x}^2=0\Rightarrow lo{g}_{-7}49=x49=(-3{)}^2$
$lo{g}_{a}b=\frac{1}{lo{g}_{b}a}$

$lo{g}_{a}b=K$
$b=a^k$

$\frac{1}{lo{g}_{b}a}=y$
$\Rightarrow lo{g}_b$
$a=\frac{1}{y}$
$a=b^{\frac{-1}{y}}=a^{\frac{x}{y}}$
$\therefore x=\frac{k}{y}\Rightarrow k=y$

Ques $\to log\left(\frac{a-b}{2}\right)=\frac{1}{2}(loga+logb)$
$a^2+b^2=Kab$ Find K
$\frac{a-b}{2}=2\sqrt{ab}\Rightarrow a^2+b^2-2ab-4ab=0a^2+b^2=6abk=6$
Indices
The product of m factors each equal to 'a' is represented by a^m
So, $a^m=a.a.a\dots \dots mtimes$
Here, a = base, m = index
Law of indices $\to 1)a^{m+n}=a^m.a^{n}2)a^{-m}=\frac{1}{a^m}a\ne 03)a^0=1a\ne 04)a^{mn}=\frac{a^m}{a^n}a\ne 05\left)\right(a^m{)}^n=a^{mn}6)a^{\frac{p}{q}}=\sqrt[q]{q{a}^p}7)(ab{)}^m=a^m.b^m$
$lo{g}_{b}a=\frac{lo{g}_{m}a}{lo{g}_{m}b}$
$a^{lo{g}_{a}N}=N=N^{lo{g}_{a}a}=Na\ne 1$
Proof: $lo{g}_{a}N=k$
$lo{g}_a\left(a^{lo{g}_{a}N}\right)=lo{g}_{a}Klo{g}_{a}N-1=lo{g}_{a}K\therefore N=K$

Antilog
If $lo{g}_{2}572=9\to 2^9=572$
$Antilo{g}_{2}9=572Antilo{g}_{2}9=2^9$
$lo{g}_{N^1}N=-1$ $lo{g}_{b^m}a=\frac{1}{m}lo{g}_{b}a$
Whomever the no. and base are on the same side of unity then logarithm of that no. to the base is +ve
When no. and base are on different sides of unity then logarithm of that no. to the base is -ve
$\to lo{g}_{10}100=2Butlo{g}_{\frac{1}{10}}100=-2$
Ques: $lo{g}_5\sqrt{5\sqrt{5\sqrt{5\sqrt{5\dots \dots }}}}=xn=?\to \left(1\right)$
$\sqrt{5\sqrt{5\sqrt{5\sqrt{5\sqrt{5}\dots \dots }}}}=y\therefore \sqrt{5y}=y\to 5y=y^2\therefore lo{g}_{5}5=1\Rightarrow y^2-5y=0y\left(y_1\right)=0y=0,5$