Question

Prove that
logab * logbc= logac

Answer 1

To prove this identity, use change of base formula:
$log_b (x) = \frac{ln x}{ln b}$

Apply to left side:
$(\frac{ln b}{ln a})*(\frac{ln c}{ln b})$

Multiply, Cancel the ln(b) factors
$= \frac{ln c}{ln a}$

Rewrite in log form:
$= log_a (c)$

This equals Right side and identity is verified.

Answer 2

Answer:

Prove that

$log_{a}(b) \times log_{b} (c)=log_{a} (c)$

To demonstarte this, we need to use the following property

$log_{b}(x)=\frac{log_{d}(x) }{log_{d}(b) }$

So, in this case,

$log_{b}(c)=\frac{log_{a}(c) }{log_{a}(b) }$

Repling this in the first expression, we have

$log_{a}(b) \times\frac{log_{a}(c) }{log_{a}(b) }=log_{a} (c)$

Multiplying and dividing, we have

$\frac{log_{a}(b) \times log_{a}(c) }{log_{a}(b) }=log_{a}(c)$

$\therefore log_{a}(c)= log_{a}(c)$

So, by using one property, we can demostrate the given expression.