Logarithms

Because 'log' kya sochenge...
Log
Log

Properties of Logarithms

1. Def:logba\log_ba is read as "log a to the base b", or in short as "log a base b" where a is the argument and b is the base

(base)(power)=(argument)=log(base)(argument)(base)^{(power)}=(argument) \;\;\;\; =\;\;\;\; \log_{(base)}(argument)

2. logba\log_{b}a is defined iff a>0,b>0,b1a > 0, b > 0, b \neq 1

iff means 'if and only if' in standard mathematics

Note that b>0,b1b>0, b \neq 1cannot be written as b>1b>1, as bbcan still be between 1 and 0

Why can't base be 1? In essence, Log finds the power given the value and base. If base is 1, value cannot be anything other than 1, as 1anything1^{anything}is still 1.

3. logaa=1\log_aa=1

Proof: Anything raised to 1 is the same thing.

4. loga1=0\log_a1 =0

Proof: If any base raised to some power is 1, then the power must be 0

5. logax+logay=loga(xy)\log_ax+\log_ay=\log_a(xy)

Proof:

logax=mam=xlogay=nan=y\log_ax=m \longrightarrow a^m = x \newline \log_ay=n \longrightarrow a^n = y \newline

Let's multiply the equations

am×an=xya^m\times a^n=xy

We can do this because we're still multiplying both sides by the same value (LHS by ana^nand RHS by yy), as ana^nis equal to yy, i.e. the same as yy

am+n=xylog(a)(xy)=m+nloga(xy)=logax+logaya^{m+n}=xy \longrightarrow \log_{(a)}(xy)=m+n \newline \therefore \log_a(xy)=\log_ax+\log_ay

6. logaxlogay=loga(xy)\log_ax - \log_ay = \log_a(\frac{x}{y})

Proof:

logax=mam=xlogay=nan=y\log_ax=m \longrightarrow a^m=x \newline \log_ay=n \longrightarrow a^n=y

Divide the equations

aman=xy\frac{a^m}{a^n} = \frac{x}{y} \newline

Which is

a(mn)=xyloga(xy)=mnloga(xy)=logaxlogaya^{(m-n)} = \frac{x}{y} \longrightarrow \log_a(\frac{x}{y}) = m-n \newline \therefore \log_a(\frac{x}{y}) = \log_ax - \log_ay

7. alogab=ba^{\log_ab} = b

Proof: Let the answer be x

alogab=xa^{\log_ab} = x

Can be written as

logax=logab\log_ax=\log_ab

...using property 1. Since bases are the same powers can be equated

x=b\therefore x=b

8. logabk=k×logab\log_a{b^k} = k \times \log_ab

Proof:

Let m=logabm=\log_ab

bm=ab^m=a

...using property 7.

(bm)k=(a)k(b^m)^k=(a)^k

Which is

bm×k=akb^{m\times k} = a^k

Using property 1...

logabk=m×k\log_ab^{k}=m\times k

As m=logabm=\log_ab

logabk=k×logab\log_a{b^k} = k \times \log_ab

9. logcalogcb=logba\frac{\log_ca}{\log_cb} = \log_ba

Proof:

Let logba=x\log_ba=x. Using property 7...

bx=ab^x = a

Take logc\log_con both sides

logcbx=logca\log_c{b^x} = \log_ca

Using property 8...

x×logcb=logcax \times \log_cb=\log_ca

Divide both sides by logcb\log_cb

x=logcalogcbx=\frac{\log_ca}{\log_cb}

Since x=logbax=\log_ba

logcalogcb=logba\frac{\log_ca}{\log_cb}= \log_ba

9.1 logba=1logab,ifc=a\log_ba = \frac{1}{log_ab}, \text{if} \; c=a

Proof:

Using property 9,

logcalogcb=logba\frac{\log_ca}{\log_cb}=\log_ba

Since in this case c=ac=a

logaalogab=logba\frac{\log_aa}{\log_ab}=\log_ba

Using property 7,

1logab=logba\frac{1}{\log_ab}=\log_ba

10. logbka=1k×logba\log_{b^k}a = \frac{1}{k} \times \log_ba

Proof:

Unwrap using property 9...

logbka=logcalogcbk\log_{b^k}a=\frac{\log_ca}{\log_cb^k}

Extract kk

=logcak×logcb= \frac{\log_ca}{k \times \log_cb}

Split the fraction

=1k×logcalogcb= \frac{1}{k} \times \frac{\log_ca}{\log_cb}

Re-wrap using property 9...

logbka=1k×logba\log_{b^k}a=\frac{1}{k} \times \log_ba

11. logbnam=mn×logba\log_{b^n}{a^m}=\frac{m}{n} \times\log_ba

This property compiles properties 9 and 10.

12. alogcb=blogcaa^{\log_cb}=b^{\log_ca}

Proof: Advanced, I may add it later

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