Question

Show that log ab×log ba =1​

Answer 1

See below for proof.

Explanation:

Given:

`\log_ab * \log_ba=1`

To prove the given equation, begin by changing the base of the two logs so that they have the same base.

`\boxed{\textsf{Change of base}: \quad \log_pr=(\log_xr)/(\log_xp)}`

Change the base of both logs to x by using the change of base formula:

`\implies \log_ab=(\log_xb)/(\log_xa)`

`\implies \log_ba=(\log_xa)/(\log_xb)`

Substitute into the equation:

`\implies \log_ab * \log_ba=(\log_xb)/(\log_xa) * (\log_xa)/(\log_xb)`

`\textsf{Apply the fraction rule} \quad (a)/(c)* (b)/(d)=(a * b)/(c * d):`

`\implies (\log_xb * \log_xa)/(\log_xa * \log_xb)`

Apply the commutative property of multiplication to the numerator:

`\implies (\log_xa * \log_xb)/(\log_xa * \log_xb)`

Cancel the common factor logₓb:

`\implies (\log_xa)/(\log_xa)`

`\textsf{Apply the fraction rule} \quad (n)/(n)=1:`

`\implies (\log_xa)/(\log_xa)=1`

Hence proving that:

• 
  `\log_ab * \log_ba=1`

In one calculation:

`\begin{aligned}\implies \log_ab * \log_ba & =(\log_xb)/(\log_xa) * (\log_xa)/(\log_xb)\\\\&=(\log_xb * \log_xa)/(\log_xa * \log_xb)\\\\&=(\log_xa * \log_xb)/(\log_xa * \log_xb)\\\\&=(\log_xa)/(\log_xa)\\\\&=1\end{aligned}`