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Prove quotient law of logrithem​

Answer 1

Answer:

For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine the quotient of exponents by subtracting:

$\sf \: xab=xa−b x a b = x a − b .$

The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms.

$\sf \: Quotient Rule: \log \: left(\dfrac{M}{N})$

(N)log, start base, b, end base, left parenthesis, start fraction, M, divided by, N, end fraction, right parenthesis, equals, log, start base, b, end base, left parenthesis, M, right parenthesis, minus, log, start base, b, end base, left parenthesis, N, right parenthesis

The proof of this property follows a method similar to the one used above.

Again, if we let M=b^xM=b x

M, equals, b, start superscript, x, end superscript and N=b^yN=b y

N, equals, b, start superscript, y, end superscript, then it follows that \log_b(M)=xlog b

(M)=xlog, start base, b, end base, left parenthesis, M, right parenthesis, equals, x and \log_b(N)=ylog b

(N)=ylog, start base, b, end base, left parenthesis, N, right parenthesis, equals,y.

We can now prove the quotient rule as follows:

$\begin{aligned}\log_b\left(\dfrac{M}{N}\right)&=\log_b\left(\dfrac{b^x}{ b^y}\right)&&\small{\gray{\text{Substitution}}}\\ \\ &=\log_b(b^{x-y})&&\small{\gray{\text{Properties of exponents}}}\\ \\ &=x-y&&\small{\gray{\log_b(b^c)=c}}\\ \\ &=\log_b(M)-\log_b(N)&&\small{\gray{\text{Substitution}}} \end{aligned}$

Hence Vertified