Stating logarithm in variables

[songoku]

Homework Statement

Given that $a=\log 3$ and $b=\log 5$, state $\log \sqrt{1000}$ in terms of u and/or v

Relevant Equations

Logarithm properties

Have tried to do that but getting no result.

I know $\log \sqrt{1000} = \frac {3}{2}$ . I just want to know whether it is possible to state $\log \sqrt{1000}$ in terms of u and/or v without using "weird stuff", like $\log \sqrt{1000} = \frac{3}{2} + u - u$ (this is what I did...)

Thanks

 

[Mark44]

Homework Statement:: Given that $a=\log 3$ and $b=\log 5$, state $\log \sqrt{1000}$ in terms of u and/or v
Relevant Equations:: Logarithm properties

Have tried to do that but getting no result.

I know $\log \sqrt{1000} = \frac {3}{2}$ . I just want to know whether it is possible to state $\log \sqrt{1000}$ in terms of u and/or v without using "weird stuff", like $\log \sqrt{1000} = \frac{3}{2} + u - u$ (this is what I did...)

Thanks

Should this have been "state $\log \sqrt{1000}$ in terms of a and/or b"? If not, I have no idea what this problem is asking.

 

[songoku]

Should this have been "state log⁡1000 in terms of a and/or b"? If not, I have no idea what this problem is asking.

Ah I am sorry, my bad. I got mixed up with another question I did.

Yes, the question should be: "state $\log⁡ \sqrt{1000}$ in terms of a and/or b"

What I did was trying something and ended up with: $\frac{3}{2} +a - a$

 

[Mark44]

Ah I am sorry, my bad. I got mixed up with another question I did.

Yes, the question should be: "state $\log⁡ \sqrt{1000}$ in terms of a and/or b"

What I did was trying something and ended up with: $\frac{3}{2} +a - a$

I don't think that's what they were looking for.

10 = 2 * 5, so $\log 10 = \log(2 * 5) = \log 2 + \log 5$
Can you work with that?

 

[songoku]

I don't think that's what they were looking for.

10 = 2 * 5, so $\log 10 = \log(2 * 5) = \log 2 + \log 5$
Can you work with that?

I can try

$\log \sqrt{1000} = \frac{3}{2} \log 10 = \frac{3}{2} (\log 5 + \log 2) = \frac{3}{2} = (b + \log 2)$

How to relate $\log 2$ to $a$ or $b$ ?

Thanks

 

[Mark44]

I can try

$\log \sqrt{1000} = \frac{3}{2} \log 10 = \frac{3}{2} (\log 5 + \log 2) = \frac{3}{2} = (b + \log 2)$

That last part should be $\frac{3}{2} (b + \log 2)$, not $\frac{3}{2} = (b + \log 2)$

How to relate $\log 2$ to $a$ or $b$ ?

2 = 10/5, so $\log 2 = \log 10 - \log 5$

 

[songoku]

That last part should be $\frac{3}{2} (b + \log 2)$, not $\frac{3}{2} = (b + \log 2)$

Yes, sorry that's typo

2 = 10/5, so $\log 2 = \log 10 - \log 5$

So:

$\log \sqrt{1000} = \frac{3}{2} \log 10 = \frac{3}{2} (\log 5 + \log 2) = \frac{3}{2} (b + \log 2)$

$=\frac{3}{2} (b + \log 10 - \log 5)$

$=\frac{3}{2} (b + 1 - b)$

$=\frac{3}{2} \rightarrow$ no $a$ or $b$ ?

Thanks

 

[Office_Shredder]

Just checking - are you sure log here is log base 10, and not natural log?

I also feel like they must have meant to say a=log(2).

The whole question as you've stated it is a bit weird.

 

[songoku]

Just checking - are you sure log here is log base 10, and not natural log?

I also feel like they must have meant to say a=log(2).

The whole question as you've stated it is a bit weird.

Yes the base of the logarithm is 10

a is log (3) because it is also used in other part of the question

Maybe the question is wrong

Thank you very much for all the help Mark44 and Office_Shredder

 

[Office_Shredder]

That would make a bit more sense, if there were eight questions all using the same a and b, and one of them happened to not need them to expressed as a fraction.

 

[songoku]

That would make a bit more sense, if there were eight questions all using the same a and b, and one of them happened to not need them to expressed as a fraction.

Unfortunately, 3/2 is stated as wrong answer by the system

 

[Office_Shredder]

Hmm. Have you tried 1.5?

I really don't know what they are looking for, sorry.

 

[archaic]

I'm not sure if this is the goal but ...
I removed the log by raising 10 to the powers $a$ and $b$, then multiplied the result$$10^a10^b=15=15\frac{1000}{1000}\Leftrightarrow\frac{10^3}{15}10^{a+b}=1000\Leftrightarrow\sqrt{\frac{10^{a+b+3}}{15}}=\sqrt{1000}$$now we just use $\log$, i guess.

 

[FactChecker]

Yes the base of the logarithm is 10

a is log (3) because it is also used in other part of the question

It sounds like there is more to the problem statement than the OP indicates. The other parts might clarify it.

 

[vela]

So:
$\log \sqrt{1000} = \frac{3}{2} \log 10 = \frac{3}{2} (\log 5 + \log 2) = \frac{3}{2} (b + \log 2)$
$=\frac{3}{2} (b + \log 10 - \log 5)$
$=\frac{3}{2} (b + 1 - b)$
$=\frac{3}{2} \rightarrow$ no $a$ or $b$ ?

Perhaps $\frac 32 = \frac {15}{10} = \frac{3\cdot 5}{10}$ can help.

 

[SammyS]

@songoku ,

Combine the idea by @archaic in Post #13: $\quad a=\log_{10}(3) \Leftrightarrow 10^a=3$ etc.

with the suggestion by @vela in Post #15.

 

[songoku]

The teacher admitted that the question is wrong.

Sorry everyone for all the trouble

 

[SammyS]

The teacher admitted that the question is wrong.

Sorry everyone for all the trouble

However, the answer you arrived at can be expressed in terms of a and b.

$\dfrac 3 2 = 10^{(a+b-1)}$

Added in Edit:
The above was corrected per post #19.

 

[songoku]

However, the answer you arrived at can be expressed in terms of a and b.

$\dfrac 2 3 = 10^{(a+b-1)}$

Yes I tried that but the correct answer came out as: $\frac{3}{2} (a+b)$ , that's why the teacher canceled the question

And I think there is typo, $\frac{2}{3}$ should be $\frac{3}{2}$

Thank you SammyS

 

[epenguin]

I get lost in long threads like this, not even sure if you have solved, but sole factors of 1000 are powers of 5 and 2. You are essentially given 5 and 2 to play with, while the problem involves 3 and it looks like you have difficulty in discovering a relation between these numbers.