Stuck on some log proofs

  • Thread starter t_n_p
  • Start date
  • #1
595
0
1) logba + logcb + logac = 1/logab + 1/logbc + 1/logca

2) logrp = q and logqr = p, show logqp = pq

3) if u = log9x, find in terms of u, logx81

4) log5x = 16logx5, solve for x

attempt

I know the change of base formula logax = logbx/logba, but I'm not sure if/how to apply it in any of the questions.

If there a trick I'm missing?
 

Answers and Replies

  • #2
eumyang
Homework Helper
1,347
10
1) logba + logcb + logac = 1/logab + 1/logbc + 1/logca
Here's a hint: looking at the first fraction,
[tex]\frac{1}{\log_a b} = \frac{\log_a a}{\log_a b}[/tex]

Can you figure out the rest, using the change of base formula?

2) logrp = q and logqr = p, show logqp = pq
Use the change of base formula for logrp to change to base q. Then substitute.

3) if u = log9x, find in terms of u, logx81
Use the change of base formula for logx81 to change to base 9.
 
Last edited:
  • #3
595
0
ok thanks, that was very helpful.
Not sure why I didn't think of this earlier, I guess I was thrown off by changing the base to a number that was already there.

I managed to do 1-3 easy, but still stuck on 4.
here is what I did:

change of base:

log5x = logxx/logx5

therefore

logxx/logx5 = 16logx5

1 = (16logx5)(logx5)

what happens with the multiplication here??
 
  • #4
eumyang
Homework Helper
1,347
10
ok thanks, that was very helpful.
Not sure why I didn't think of this earlier, I guess I was thrown off by changing the base to a number that was already there.

I managed to do 1-3 easy, but still stuck on 4.
here is what I did:

change of base:

log5x = logxx/logx5

therefore

logxx/logx5 = 16logx5

1 = (16logx5)(logx5)

what happens with the multiplication here??
I would use the change of base with logx5 to change to base 5 instead. At some point you will need to take the square root of both sides and use the basic definition
[tex]\log_b a = y \leftrightarrow b^y = a[/tex] to find your answer.
 
  • #5
595
0
ok, when I take square root, I get two solutions, from the + -.

I get x = 625 and x = 1/625

the answer only lists x=625, I can't see why x=1/625 is invalid.

any ideas?
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
...

the answer only lists x=625, I can't see why x=1/625 is invalid.

any ideas?
Plug in 1/625 to check.

[tex]5^{-4}=1/625\ \to\ \log_5 \left(\frac{1}{625}\right)=-4\ .[/tex]

Now see if the right hand side is -4.
 
  • #7
595
0
when you say check the right hand side, are you referring to this equation:

log5x = 16logx5?

i.e. check that 16log1/6255 = -4?

I checked this like so:

16log1/6255 = 16log1/625(1/625)-1/4
= -16/4
= -4

I don't understand the logic in doing this.
This is where I started and got my answer from, so when I put it back its going to give me -4, like a chicken and the egg problem.

Have I lost the plot?
 
  • #8
eumyang
Homework Helper
1,347
10
ok, when I take square root, I get two solutions, from the + -.

I get x = 625 and x = 1/625

the answer only lists x=625, I can't see why x=1/625 is invalid.

any ideas?
Typo, maybe? Or was there a restriction for x that was not stated? Without any restrictions, x = 1/625 is valid.

I checked this like so:

16log1/6255 = 16log1/625(1/625)-1/4
= -16/4
= -4

I don't understand the logic in doing this.
This is where I started and got my answer from, so when I put it back its going to give me -4, like a chicken and the egg problem.
What is it that you don't understand?
 
  • #9
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
when you say check the right hand side, are you referring to this equation:

log5x = 16logx5?

i.e. check that 16log1/6255 = -4?

I checked this like so:

16log1/6255 = 16log1/625(1/625)-1/4
= -16/4
= -4

I don't understand the logic in doing this.
This is where I started and got my answer from, so when I put it back its going to give me -4, like a chicken and the egg problem.

Have I lost the plot?
It's no more a circular argument than when you check a result after solving any other equation in algebra.

It looks like x = 1/625 is as good an answer as x = 625 .

I suspect the reason that the only answer listed was x=625, is that it's unusual to use a number smaller than 1 as the base for a logarithm.
 
Last edited:
  • #10
595
0
It's no more a circular argument than when you check a result after solving any other equation in algebra.

It looks like x = 1/625 is as good an answer as x = 625 .

I suspect the reason that the only answer listed was x=625, is that it's unusual to use a number smaller than 1 as the base for a logarithm.
Regarding the check, you are right. When you said check I thought you meant check as a valid/invalid thing as opposed to just a simply algebra check. For example, if x = -5, when I put it back then the answer would be invalid as I can't take log of a negative number.

There was no restriction in the question, so I'm happy to list the two answers. You say its unusual to see a number smaller than 1 as the base in a log, but that doesn't really take away from the fact it should still be listed. Would you agree?

Thanks to all for the help
 
  • #11
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
Yes, I agree. They are both solutions.
 

Related Threads on Stuck on some log proofs

  • Last Post
Replies
15
Views
1K
  • Last Post
Replies
1
Views
914
Replies
8
Views
909
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
614
Replies
16
Views
858
  • Last Post
Replies
5
Views
888
  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
15
Views
1K
Top