Question

Prove that
logab * logbc= logac

Answer 1

To prove this identity, use change of base formula:

`log_b (x) = (ln x)/(ln b)`

Apply to left side:

`((ln b)/(ln a))*((ln c)/(ln b))`

Multiply, Cancel the ln(b) factors

`= (ln c)/(ln a)`

Rewrite in log form:

`= log_a (c)`

This equals Right side and identity is verified.

Answer 2

Prove that

`log_(a)(b) * log_(b) (c)=log_(a) (c)`

To demonstarte this, we need to use the following property

`log_(b)(x)=(log_(d)(x) )/(log_(d)(b) )`

So, in this case,

`log_(b)(c)=(log_(a)(c) )/(log_(a)(b) )`

Repling this in the first expression, we have

`log_(a)(b) *(log_(a)(c) )/(log_(a)(b) )=log_(a) (c)`

Multiplying and dividing, we have

`(log_(a)(b) * 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.