Question

Prove the logarithmic property?

Let `b, x, y in RR^+` such that `b!= 1`
Prove that:
`log_(1/b)x=-log _b x`
I have tried doing it quite a few times each time using a different method, but I still do something wrong. I think I am just making a small mistake somewhere but I have run out of ideas on how to tackle this one.
Thank you very much.

Answer 1

Please refer to a Proof in Explanation.

Explanation:

Suppose that, `log_(1/b)x=m`.

`:. (1/b)^m=x`.

`:. (b^-1)^m=x`.

`:. b^-m=x`.

`:. log_b(b^-m)=log_bx`.

`:. -mlog_b b=log_bx`.

`:. -m*1=log_b x`.

`:. m=-log_b x`.

`or, log_(1/b)x=-log_b x`, as desired!

Answer 2

Please see below.

Explanation:

We know that , for `m,n inRR^+ and k in RR^(+) -{1}`

`color(red)((1)"Change of Base formula :"`

`color(red)(log_n m=(log_k m)/(log_k n`

`color(blue)((2)"Quotient/Difference formula :"`

`color(blue)(log_k(A/B)=log_k A-log_kB`

We take LHS :

`LHS=color(red)(log_(1/b) x)............tocolor(red)(Apply(1)`

`LHS=color(red)((log_k x)/(log_k (1/b)`

Using `color(blue)((2)` in the denominator ,we get

`LHS=(log_k x)/color(blue)((log_k 1-log_k b))`

`LHS=(log_k x)/(0-log_k b)to[because log_k1=0]`

`LHS=-log_kx /log_k b`

Again using `(1)`

`LHS==-log_b x`

`LHS=RHS`