# Stuck on some log proofs

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?

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eumyang
Homework Helper
1) logba + logcb + logac = 1/logab + 1/logbc + 1/logca
Here's a hint: looking at the first fraction,
$$\frac{1}{\log_a b} = \frac{\log_a a}{\log_a b}$$

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.

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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??

eumyang
Homework Helper
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
$$\log_b a = y \leftrightarrow b^y = a$$ to find your answer.

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?

SammyS
Staff Emeritus
Homework Helper
Gold Member
...

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

any ideas?
Plug in 1/625 to check.

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

Now see if the right hand side is -4.

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?

eumyang
Homework Helper
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?

SammyS
Staff Emeritus
Homework Helper
Gold Member
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.

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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

SammyS
Staff Emeritus