How to Solve Logarithmic Equations Using Change of Base Formula?

[t_n_p]

1) log_ba + log_cb + log_ac = 1/log_ab + 1/log_bc + 1/log_ca

2) log_rp = q and log_qr = p, show log_qp = pq

3) if u = log₉x, find in terms of u, log_x81

4) log₅x = 16log_x5, solve for x

attempt

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

If there a trick I'm missing?

 

[eumyang]

1) log_ba + log_cb + log_ac = 1/log_ab + 1/log_bc + 1/log_ca

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) log_rp = q and log_qr = p, show log_qp = pq

Use the change of base formula for log_rp to change to base q. Then substitute.

3) if u = log₉x, find in terms of u, log_x81

Use the change of base formula for log_x81 to change to base 9.

 

[t_n_p]

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 = log_xx/log_x5

therefore

log_xx/log_x5 = 16log_x5

1 = (16log_x5)(log_x5)

what happens with the multiplication here??

 

[eumyang]

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 = log_xx/log_x5

therefore

log_xx/log_x5 = 16log_x5

1 = (16log_x5)(log_x5)

what happens with the multiplication here??

I would use the change of base with log_x5 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.

 

[t_n_p]

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]

...

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.

 

[t_n_p]

when you say check the right hand side, are you referring to this equation:

log₅x = 16log_x5?

i.e. check that 16log_1/6255 = -4?

I checked this like so:

16log_1/6255 = 16log_1/625(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]

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:

16log_1/6255 = 16log_1/625(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]

when you say check the right hand side, are you referring to this equation:

log₅x = 16log_x5?

i.e. check that 16log_1/6255 = -4?

I checked this like so:

16log_1/6255 = 16log_1/625(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.

 

[t_n_p]

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]

Yes, I agree. They are both solutions.