SOLUTION: Hi Tutor, Can you please explain to me why log(base b)x/log(base b)y does not equal log(base b)x

SOLUTION: Hi Tutor, Can you please explain to me why log(base b)x/log(base b)y does not equal log(base b)x - log(base b)y? Thank you

Question 274525: Hi Tutor,
Can you please explain to me why log(base b)x/log(base b)y does not equal log(base b)x - log(base b)y? Thank you
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
I will use the base of 10 to show you because that's what the calculator uses.

The answer applies to all bases.

The rule states that:

log(x/y) = log(x) - log(y)

The rule does not state that:

log(x)/log(y) = log(x) - log(y)

your second statement is clearly invalid if you look at it the following way.

let a = log(x)
let b = log(y)

log(x)/log(y) = log(x) - log(y) becomes:

a/b = a-b which is clearly not valid.

if we use numbers, you will see that the stated rule is valid while the alternate rule you are asking about is not.

let x = 1000
let y = 10
let x/y = 1000/10 = 100

use your calculator to derive:

log(1000) = 3
log(10) = 1
log(1000/10) = log(100) = 2

the rule states that:

log(x/y) = log(x) - log(y)

putting it into numbers, this means that:

log(1000/10) = log(100) = log(1000) - log(10)

since log(100) = 2 and log(1000) = 3 and log(10) = 1, this equation becomes:

2 = 3-1 which is true.

the rule works.

use the alternate rule and you have:

log(x)/log(y) = log(x) - log(y) which becomes:

log(1000) / log(10) = log(1000) - log(10) which becomes:

3/1 = 3-1 which is false.

the alternate rule you asked about does not work.

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