Convert log(a/b) to - log(b/a), show the steps?

Question

Here is my answer, is this right? If not, can anyone show me how to do it the right way? Thanks!!

Since a/b= a(1/b)= a*b^(-1),

and we know that log (a^x)= x*loga

Therefore, Log(a/b)=log(a*b^-1)=log a+logb^-1

=-log b+log a= -(logb-log a)= - log(b/a)

Answer 1

Not at all wrong, just a little bit the hard way.

Quotient rule says log(a/b) = log a - log b, so

log(a/b) = log a - log b

log(a/b) = - [ - (log a - log b)] .......... double negative

log(a/b) = - [ log b - log a] ............... distributive

log(a/b) = - log(b/a) ........................ quotient rule again.

Answer 2

Rule:

Log (a/b) = Log(a) - Log(b)

Proof:

Let x = Log(a), y = Log(b)

then a = 10^x, b = 10^y

a/b = 10^x / 10^y = 10^x 10^(-y)

or  a/b = 10^(x - y)

so Log (a/b) = x - y = Log(a) - Log(b)

& Log (b/a) = Log(b) - Log(a) = - Log (a/b)

  

Answer 3

The rule is  log(a/b)= loga - logb=-(logb - loga)=-log(b/a)