Does this look correct?

[allegansveritatem]

I am sorry I don't have a copy of the problem but it is pretty simple and I think what I present here makes it clear. I am being asked to write the first formula for L in terms of m and Lo. Is what I have done the proper procedure here? It seems to work but looks odd to me for some reason:

 

[tkhunny]

The right hand side of that third line is utterly senseless. I realize it's a map of where you are going, but just don't ever write anything like that.

Are you SURE the original logarithm is Base 10?

Please simplify, especially negative signs. It is very easy to make mistakes with something like [math]\dfrac{M-6}{-2.5}[/math]. Please consider writing [math]\dfrac{6-M}{2.5}[/math]. There's a whole lot less to go wrong.

 

[topsquark]

Looks good to me. I'd get rid of the negative in the fraction in the exponent, though.

Another thought. I don't think it's needed but you could also separate out the m in the exponent:
[math]10^{(m - 6)/(-2.5)} = 10^{m/(-2.5)} \cdot 10^{-6/(-2.5)} = 251.19 \cdot 10^{-m/2.5}[/math]which gives
[math]L = \left ( 251.19 L_0 \right ) \cdot 10^{-m/2.5}[/math]
-Dan

 

[Dr.Peterson]

The work and answer are right, except that the RHS of the third line should be [MATH]10^{\log_{10}\left(\frac{L}{L_0}\right)}[/MATH]. You didn't write what you meant.

 

[allegansveritatem]

The right hand side of that third line is utterly senseless. I realize it's a map of where you are going, but just don't ever write anything like that.

Are you SURE the original logarithm is Base 10?

Please simplify, especially negative signs. It is very easy to make mistakes with something like [math]\dfrac{M-6}{-2.5}[/math]. Please consider writing [math]\dfrac{6-M}{2.5}[/math]. There's a whole lot less to go wrong.

As far as I can make out it was base 10 in that no base was specified. Thanks for the tip re: the negative signs. I will remember that.

 

[allegansveritatem]

Looks good to me. I'd get rid of the negative in the fraction in the exponent, though.

Another thought. I don't think it's needed but you could also separate out the m in the exponent:
[math]10^{(m - 6)/(-2.5)} = 10^{m/(-2.5)} \cdot 10^{-6/(-2.5)} = 251.19 \cdot 10^{-m/2.5}[/math]which gives
[math]L = \left ( 251.19 L_0 \right ) \cdot 10^{-m/2.5}[/math]
-Dan

good idea.

 

[allegansveritatem]

The work and answer are right, except that the RHS of the third line should be [MATH]10^{\log_{10}\left(\frac{L}{L_0}\right)}[/MATH]. You didn't write what you meant.

Right, I meant to raise 10 to the right side and 10 to the left. That log 10 below is a zombie product.

 

[Steven G]

Do not assume that if there is no base listed then the base is 10. The base can be e.
How do you tell? The textbook or instructor has to state this somewhere. To be clear, some books/teachers use Log x to mean Log₁₀ while other books/teachers use Log x to mean ln x.

 

[Dr.Peterson]

As far as I can make out it was base 10 in that no base was specified.

You can "assume" that "log" means base 10, based on the conventions of your textbook. That's not really a mere assumption, so it's fine. But it's a reason to state what conventions you are using, or for us to ask in order to be sure.

 

[JeffM]

In calculus, according to some,

[MATH]log(x) \equiv log_e(x),[/MATH] which is definitely convenient in calculus.

In other areas, log to the base 10 is often convenient and thus

[MATH]log(x) \equiv log_{10}(x) \text { and } log_e(x) = l n(x).[/MATH]
In computer science, frequently what is convenient is

[MATH]log(x) \equiv log_2(x).[/MATH]
But the line immediately preceding is restricted pretty much exclusively to those who worry about bits, bytes, and CPU times.

Personally, I like the old-fashioned [MATH]log(x) \equiv log_{10}(x) \text { and } ln(x) = log_e(x).[/MATH]

 

[allegansveritatem]

In calculus, according to some,

[MATH]log(x) \equiv log_e(x),[/MATH] which is definitely convenient in calculus.

In other areas, log to the base 10 is often convenient and thus

[MATH]log(x) \equiv log_{10}(x) \text { and } log_e(x) = l n(x).[/MATH]
In computer science, frequently what is convenient is

[MATH]log(x) \equiv log_2(x).[/MATH]
But the line immediately preceding is restricted pretty much exclusively to those who worry about bits, bytes, and CPU times.

Personally, I like the old-fashioned [MATH]log(x) \equiv log_{10}(x) \text { and } ln(x) = log_e(x).[/MATH]

the book I have would be to your taste then: log x means log base10 x and ln x means log base e x. Good enough for me.