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.

2 Answers
Jul 15, 2018

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!

Jul 15, 2018

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