Respuesta :
Answer:
Step-by-step explanation:
Given expression = [tex]log_b\frac{x}{y}[/tex]
To Prove : a) [tex]log_b\frac{x}{y}=log_bx+log_by^{-1} [/tex]
b)[tex]log_b\frac{x}{y}=log_bx-log_by[/tex]
c) [tex]log_b\frac{x}{y}=log_bxy^{-1} [/tex]
d) [tex]log_b\frac{x}{y}=log_bx-1log_by[/tex]
Solution :
a) [tex]log_b\frac{x}{y}=log_bx+log_by^{-1} [/tex]
[tex]LHS = log_b\frac{x}{y}[/tex]
Using identity [tex]\frac{1}{y} = y^{-1}[/tex]
⇒ [tex]LHS = log_bxy^{-1}[/tex]
Using identity : [tex]log_bmn = log_bm +log_bn[/tex]
⇒ [tex]LHS =log_bx+log_by^{-1}[/tex]
Hence LHS = RHS
Hence proved.
b) [tex]log_b\frac{x}{y}=log_bx-log_by[/tex]
[tex]LHS = log_b\frac{x}{y}[/tex]
Using identity : [tex]log_b\frac{m}{n} = log_bm -log_bn[/tex]
⇒ [tex] LHS=log_bx-log_by[/tex]
LHS= RHS
Hence proved.
c)[tex]log_b\frac{x}{y}=log_bxy^{-1} [/tex]
[tex]LHS = log_b\frac{x}{y}[/tex]
Using identity [tex]\frac{1}{y} = y^{-1}[/tex]
⇒ [tex]LHS = log_bxy^{-1} [/tex]
LHS=RHS
Hence proved .
d) [tex]log_b\frac{x}{y}=log_bx-1log_by[/tex]
[tex]LHS = log_b\frac{x}{y}[/tex]
Using part a
⇒ [tex]LHS =log_bx+log_by^{-1}[/tex]
Using identity : [tex]log_bm^{-1} = -1log_bm[/tex]
⇒ [tex]LHS=log_bx-1log_by[/tex]
LHS = RHS
Hence proved
