Proofs of Logarithmic properties

1) Product Rule: Log_a xy = log_a x+ log_a y

proof:

let log_a x = m, log_a y = n

convert these logarithms into exponents

x=a^m and y=aⁿ

multiply the above equations

xy=a^m * aⁿ

Using the property b^p * b^q=b^p+q from exponents we can write the above equation as follows

xy=a^m+n

Convert back into logarithm

log_a xy = m+n

on substituting the values m= log_a x and n =log_a y we get

log_a xy = log_a x + log_a y

2) Quotient Rule: Log_a $\frac{x}{y}$ =log_a x- log_a y

proof:

let log_a x = m, log_a y = n

convert these logarithms into exponents

x=a^m and y=aⁿ

divide the above equations

$\frac{x}{y}$ = $\frac{a^m}{a^n}$

using the property b^p/b^q=b^p-q from exponents we can write the above equation as follows

$\frac{x}{y}$=a^m-n

convert into logarithm

log_a $\frac{x}{y}$=m-n

on substituting the m= log_a x & n =log_a y we get

log_a (xy)=log_a x - log_a y

3) Power Rules:

i) log_b $a^m$ = m log_b a

Proof:

let x=log_b a

convert logarithm into exponent

b^x=a

taking both sides of the equation to m’th power

(b^x)^m=a^m

using the property (a^m)ⁿ=a^mn from exponents we can write the above equation as follows

b^xm=a^m

convert back to logarithm

xm=log_b a^m

substitute x=log_b a back in the equation and interchanging

log_b a^m = m log_b a

ii) log_bⁿ a= $\frac{1}{n}$ log_b a

proof:

let log_bⁿ a=x

convert this into exponential form

a=(bⁿ)^x

use exponential property

(a^m)ⁿ=a^mn=(aⁿ)^m

a=(b^x)ⁿ

use the property if a^m=b then a=b^{$\frac{1}{m}$} from exponents we can write the above equation as follows

a^{$\frac{1}{n}$}=b^x

interchanging and converting into logarithmic form

b^x=a^{$\frac{1}{n}$}

x=log^b a^{$\frac{1}{n}$}

substitute the value of x=log_bⁿ a and use logarithmic property log_b a^m= m log_b a

log_bⁿ a= $\frac{1}{n}$ log_b a

4) log of 1 rule: log_b 1=0

Proof:

from exponents we have:

b⁰=1

convert this into logarithmic form

0=_b 1

5) Change of Base Rule: log_b a= $\frac{log_x a}{log_x b}$

Proof:

let log_x a=p, log_x b=q

Convert these logarithms into exponential forms

a=x^p and b=x^q

Substituting in the Left hand side of the to solve into right hand side of the Rule

L.H.S. : log_b a= log_xⁿ x^m

Using the power rule: log_n^q a^p= $\frac{p}{q}$ $log_b$ a

log_b a= $\frac{p}{q}$ log_x x

Using the Rule log_a a=1

Log_b a= $\frac{p}{q}$

Substituting the assumed p= $log_x$ a, q= $log_x$ b

Log_b a= $\frac{log_x a}{log_x b}$

6) log_b $\frac{1}{a}$ =log_{$\frac{1}{b}$} a= -log_b a

Proof:

i) Log_b (1/a)=log_b a^-1

Using the property log_a b^m=m log_a b =-log_b a

ii) log_{$\frac{1}{b}$} a=log_$b^{-1}$ a

log_b $\frac{1}{a}$ =log_{$\frac{1}{b}$} a= -log_b a

7) a^{log_a x} = x

Proof:

Let p=a^{log_a x}

Converting the above exponent into logarithm

Log_a p = log_a x

Using he property if log_b a= log_b c then a=c

p=x

But we assumed p=a^{log_a x}

Therefore a^{log_a x} = x