# Math Help - Log proofs

1. ## Log proofs

Hi I am terrible at proofs. I am going to write 'log base b' as logb. I need to prove the following:
1. logb 1=0
2. logb b=1
3. logb 1/b=-1
4. logb 1/x=-logb x

I realise this may a bit time consuming so thank you in advance. Or if anyone has seen these proofs somewhere? Thanks.

2. Originally Posted by slaypullingcat
Hi I am terrible at proofs. I am going to write 'log base b' as logb. I need to prove the following:
1. logb 1=0
2. logb b=1
3. logb 1/b=-1
4. logb 1/x=-logb x

I realise this may a bit time consuming so thank you in advance. Or if anyone has seen these proofs somewhere? Thanks.
In each case you should use the fact that $\log_b a = c \iff b^c = a$.

3. Originally Posted by slaypullingcat
Hi I am terrible at proofs. I am going to write 'log base b' as logb. I need to prove the following:
1. logb 1=0
2. logb b=1
3. logb 1/b=-1
4. logb 1/x=-logb x

I realise this may a bit time consuming so thank you in advance. Or if anyone has seen these proofs somewhere? Thanks.
These all follow, fairly immediately, from the basic definition of logarithms, as pointed out in the previous reply. Where are you having difficulty?

4. I'm having trouble with the fourth one. I guess I can show the relationships as stated in the link you posted, but, I don't feel like I have proved anything.

For example, with No.3, I just wrote
log b (1/b)=-1 then log 4 (1/4)=-1 because 4^-1=1/4
Thats it, I did the same for the first two but I'm stuck on the fourth. I just don't know if what I have done constitutes a proof?
Thank you

5. Originally Posted by slaypullingcat
I'm having trouble with the fourth one. I guess I can show the relationships as stated in the link you posted, but, I don't feel like I have proved anything.

For example, with No.3, I just wrote
log b (1/b)=-1 then log 4 (1/4)=-1 because 4^-1=1/4
Thats it, I did the same for the first two but I'm stuck on the fourth. I just don't know if what I have done constitutes a proof?
Thank you
$\log_b \frac{1}{x} = c$ .... (1)

$\Rightarrow b^c = \frac{1}{x} \Rightarrow b^{-c} = x$

$\Rightarrow -c = \log_b x \Rightarrow c = - \log_b x$ .... (2)

Now equate (1) and (2).