# Proofs of Logarithmic properties

1) Product Rule: Loga xy = loga x+ loga y

proof:

let loga x = m, loga y = n

convert these logarithms into exponents

x=am and y=an

multiply the above equations

xy=am * an

Using the property bp * bq=bp+q from exponents we can write the above equation as follows

xy=am+n

Convert back into logarithm

loga xy = m+n

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

loga xy = loga x + loga y

2) Quotient Rule: Loga $\frac{x}{y}$ =loga x- loga y

proof:

let loga x = m, loga y = n

convert these logarithms into exponents

x=am and y=an

divide the above equations

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

using the property bp/bq=bp-q from exponents we can write the above equation as follows

$\frac{x}{y}$=am-n

convert into logarithm

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

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

loga (xy)=loga x - loga y

3) Power Rules:

i) logb $a^m$ = m logb a

Proof:

let x=logb a

convert logarithm into exponent

bx=a

taking both sides of the equation to m’th power

(bx)m=am

using the property (am)n=amn from exponents we can write the above equation as follows

bxm=am

convert back to logarithm

xm=logb am

substitute x=logb a back in the equation and interchanging

logb am = m logb a

ii) logbn a= $\frac{1}{n}$ logb a

proof:

let logbn a=x

convert this into exponential form

a=(bn)x

use exponential property

(am)n=amn=(an)m

a=(bx)n

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

a$\frac{1}{n}$=bx

interchanging and converting into logarithmic form

bx=a$\frac{1}{n}$

x=logb a$\frac{1}{n}$

substitute the value of x=logbn a and use logarithmic property logb am= m logb a

logbn a= $\frac{1}{n}$ logb a

4) log of 1 rule: logb 1=0

Proof:

from exponents we have:

b0=1

convert this into logarithmic form

0=b 1

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

Proof:

let logx a=p, logx b=q

Convert these logarithms into exponential forms

a=xp and b=xq

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

L.H.S. : logb a= logxn xm

Using the power rule: lognq ap= $\frac{p}{q}$ $log_b$ a

logb a= $\frac{p}{q}$ logx x

Using the Rule loga a=1

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

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

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

6) logb $\frac{1}{a}$ =log$\frac{1}{b}$ a= -logb a

Proof:

i) Logb (1/a)=logb a-1

Using the property loga bm=m loga b =-logb a

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

logb $\frac{1}{a}$ =log$\frac{1}{b}$ a= -logb a

7) aloga x = x

Proof:

Let p=aloga x

Converting the above exponent into logarithm

Loga p = loga x

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

p=x

But we assumed p=aloga x

Therefore aloga x = x