logarithms

[Albi]

If log(a)x=2, log(b)x=3 and log(c)x=6 then how much is log(abc)x going to be?

 

[Steven G]

If the x represents the base, then log_x(abc) = log_x(a) + log_x(b) + log_x(c)

 

[Albi]

If the x represents the base, then log_x(abc) = log_x(a) + log_x(b) + log_x(c)

No I didn't mean to write x as the base

 

[pka]

No I didn't mean to write x as the base

Well that is one way to have the question make sense.
What is(are) the base(s)?

 

[Steven G]

No I didn't mean to write x as the base

Fine, can you please tell us what you did mean?

Maybe you meant (log(a))x=2, (log(b))x=3 and (log(c)x)=6. You need to tell us what you are taking the logs of.

 

[Steven G]

11x is my guess

 

[Albi]

Fine, can you please tell us what you did mean?

Maybe you meant (log(a))x=2, (log(b))x=3 and (log(c)x)=6. You need to tell us what you are taking the logs of.

I meant log base a of x equals 2 and so on

 

[pka]

So we have \(\log_{a}(x)=2,~\log_{b}(x)=3,~\&~\log_{c}(x)=6\)
OR \(x=a^2,~x=b^3~\&~x=c^6\)
Now what?

 

[Albi]

So we have \(\log_{a}(x)=2,~\log_{b}(x)=3,~\&~\log_{c}(x)=6\)
OR \(x=a^2,~x=b^3~\&~x=c^6\)
Now what?

I don't know mabye I should try to solve the other way around because this really doesn't make sense

 

[pka]

I don't know mabye I should try to solve the other way around because this really doesn't make sense

Actually the question as written makes no sense.
Please go back nd reread the question and post it exactly as written.
The Latex code is \log_b(x)=3

 

[Dr.Peterson]

If log(a)x=2, log(b)x=3 and log(c)x=6 then how much is log(abc)x going to be?

I understand this to mean

If log_a(x)=2, log_b(x)=3, and log_c(x)=6 then how much is log_abc(x) going to be?​

When I see an unknown as the base of a log, I immediately think about the fact that log_x(y) = 1/log_y(x). Are you familiar with that? (It can be easily proved using the change of base formula.)

Use that to rewrite all four equations in the problem, and you'll find that the answer is easy!

(Using pka's idea in post #8, you can express a, b, and c in terms of x, and then express abc likewise. The answer again drops out naturally.)

 

[Cubist]

Don't we just need to extend @pka 's post#8 a bit by writing...

[math]x = a^2 = b^3 = c^6 = (abc)^y[/math] and then find y (which is the answer)

Start by writing "a" and "b" in terms of c, then you can substitute these values into [math](abc)^y = c^6[/math] to find y

 

[Albi]

I understand this to mean

If log_a(x)=2, log_b(x)=3, and log_c(x)=6 then how much is log_abc(x) going to be?​

When I see an unknown as the base of a log, I immediately think about the fact that log_x(y) = 1/log_y(x). Are you familiar with that? (It can be easily proved using the change of base formula.)

Use that to rewrite all four equations in the problem, and you'll find that the answer is easy!

(Using pka's idea in post #8, you can express a, b, and c in terms of x, and then express abc likewise. The answer again drops out naturally.)

Can you explain it a bit more cause I don't get it.

 

[Dr.Peterson]

Can you explain it a bit more cause I don't get it.

Please make an attempt and show your work. It's a lot easier to guide someone when you can see where they are not understanding.

If nothing else, tell me whether you know the change of base formula.

 

[Albi]

Please make an attempt and show your work. It's a lot easier to guide someone when you can see where they are not understanding.

If nothing else, tell me whether you know the change of base formula.

No I don't know this formula

 

[Steven G]

if x=a^2, then a=?
If x =b^3, then b =?
If x = c^6, the c =?

Now multiply a , b and c and get abc = x^? Then log_abcx = ?

 

[pka]

No I don't know this formula

I will tell you this answer if you tell us if you reviewed the original question to be clear that you posted it correctly.
If \(T=\log_b(S)\) then we can change the base to \(T=\dfrac{\log(S)}{\log(b)}\)
Example \(\log_{10}(75)=\dfrac{\log(75)}{\log(10)}\).
So what Prof Peterson means: if \(\log_x(y)=\dfrac{\log(y)}{\log(x)}=\dfrac{1}{\frac{\log(x)}{\log(y)}}=\dfrac{1}{\log_y(x)}\)

 

[Albi]

I will tell you this answer if you tell us if you reviewed the original question to be clear that you posted it correctly.
If \(T=\log_b(S)\) then we can change the base to \(T=\dfrac{\log(S)}{\log(b)}\)
Example \(\log_{10}(75)=\dfrac{\log(75)}{\log(10)}\).
So what Prof Peterson means: if \(\log_x(y)=\dfrac{\log(y)}{\log(x)}=\dfrac{1}{\frac{\log(x)}{\log(y)}}=\dfrac{1}{\log_y(x)}\)

Yeah I checked the question and it is posted correctly

 

[Dr.Peterson]

Yes, I knew it was posted correctly, because it has a very nice answer.

Now please show SOME bit of work so we can tell what you CAN do, rather than just what you can't. Try rewriting each equation in terms of a base-x log, and show us what you get. Also think about what you might be able to do with the results.

 

[Albi]

Yes, I knew it was posted correctly, because it has a very nice answer.

Now please show SOME bit of work so we can tell what you CAN do, rather than just what you can't. Try rewriting each equation in terms of a base-x log, and show us what you get. Also think about what you might be able to do with the results.

I tried to rewrite each equation as a log x base but I don't know if I have done it correctly and I don't know what to do with 2,3 and 6