Answer:
logₓ(y) = [tex]\frac{\log(y)}{\log(x)}[/tex]
[tex]\frac{\log(b)}{\log(a)}\times\frac{\log(a)}{\log(b)}[/tex]
or
= 1
Step-by-step explanation:
Data provided:
a ≠ 1 and b ≠ 1
To prove [tex]\log_a(b).\log_b(a)=1[/tex]
RHS = 1
LHS = [tex]\log_a(b).\log_b(a)[/tex]
Now,
We have the property of the log function as:
logₓ(y) = [tex]\frac{\log(y)}{\log(x)}[/tex]
applying the above property on the LHS side to solve LHS, we get
LHS = [tex]\frac{\log(b)}{\log(a)}\times\frac{\log(a)}{\log(b)}[/tex]
or
LHS = 1
Since,
LHS = 1 is equal to the RHS = 1
Hence, proved that [tex]\log_a(b).\log_b(a)=1[/tex]
