Question

Choose the letter of the expression listed on the right that completes each step to show how to use the power and product properties of logarithms to prove that the quotient property is true for logb x y . logb x y = = = = a logbx + logby-1 b logbx - logby c logbxy-1 d logbx - 1logby

Answer 1

C
A
D
B

Hope it helps :)))))))

Answer 2

Answer:

Step-by-step explanation:

Given expression = $log_b\frac{x}{y}$

To Prove : a) $log_b\frac{x}{y}=log_bx+log_by^{-1}$

                  b)$log_b\frac{x}{y}=log_bx-log_by$

                  c) $log_b\frac{x}{y}=log_bxy^{-1}$

                  d)  $log_b\frac{x}{y}=log_bx-1log_by$

Solution :

a) $log_b\frac{x}{y}=log_bx+log_by^{-1}$

$LHS = log_b\frac{x}{y}$

Using identity $\frac{1}{y} = y^{-1}$

⇒ $LHS = log_bxy^{-1}$

Using identity :  $log_bmn = log_bm +log_bn$

⇒ $LHS =log_bx+log_by^{-1}$

Hence LHS = RHS

Hence proved.

b) $log_b\frac{x}{y}=log_bx-log_by$

$LHS = log_b\frac{x}{y}$

Using identity : $log_b\frac{m}{n} = log_bm -log_bn$

⇒ $LHS=log_bx-log_by$

LHS= RHS

Hence proved.

c)$log_b\frac{x}{y}=log_bxy^{-1}$

$LHS = log_b\frac{x}{y}$

Using identity $\frac{1}{y} = y^{-1}$

⇒ $LHS = log_bxy^{-1}$

LHS=RHS

Hence proved .

d) $log_b\frac{x}{y}=log_bx-1log_by$

$LHS = log_b\frac{x}{y}$

Using part a

⇒ $LHS =log_bx+log_by^{-1}$

Using identity : $log_bm^{-1} = -1log_bm$

⇒ $LHS=log_bx-1log_by$

LHS = RHS

Hence proved