Question

Show that for any positive numbers a and b with a ≠ 1 and b ≠ 1, loga(b) ∙ logb(a) = 1.

Answer 1

Answer:

logₓ(y) = $\frac{\log(y)}{\log(x)}$

$\frac{\log(b)}{\log(a)}\times\frac{\log(a)}{\log(b)}$

or

= 1

Step-by-step explanation:

Data provided:

a ≠ 1 and b ≠ 1

To prove $\log_a(b).\log_b(a)=1$

RHS = 1

LHS = $\log_a(b).\log_b(a)$

Now,

We have the property of the log function as:

logₓ(y) = $\frac{\log(y)}{\log(x)}$

applying the above property on the LHS side to solve LHS, we get

LHS = $\frac{\log(b)}{\log(a)}\times\frac{\log(a)}{\log(b)}$

or

LHS = 1

Since,

LHS = 1 is equal to the RHS = 1

Hence, proved that  $\log_a(b).\log_b(a)=1$