Log Simplification Strategies for Evaluating Expressions with Square Roots

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MacLaddy
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Homework Statement



Assume [itex]log_bx=0.36, log_by=0.56, log_bz=0.83[/itex]

Evaluate the following expressions

[tex]log_b\frac{\sqrt{xy}}{z}[/tex]

Homework Equations





The Attempt at a Solution



I know that the first step of simplification is [itex](log_b\sqrt{x}+log_b\sqrt{y})-(log_bz)[/itex], and that [itex](log_bz) = 0.83[/itex], so that ends up being [itex](log_b\sqrt{x}+log_b\sqrt{y})-(0.83)[/itex]

However, I have no idea what to do with the [itex]\sqrt{x}[/itex] that is in the first two logs. I can't seem to find the right search online for that, and it's not in my book.

Any help would be greatly appreciated.

Thanks,
Mac
 
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I guess I should clarify what I am asking.

If [itex]log_bx = 0.36,[/itex] then does [itex]log_b\sqrt{x}=\sqrt{0.36}[/itex]
 
MacLaddy said:
I guess I should clarify what I am asking.

If [itex]log_bx = 0.36,[/itex] then does [itex]log_b\sqrt{x}=\sqrt{0.36}[/itex]
The quick answer ... No.

By the rule of logarithms. [itex]\displaystyle\log_b(\sqrt{x})=\frac{1}{2}\log_b(x)\,.[/itex]

For one thing, [itex]\sqrt{x}=x^{1/2}\,.[/itex]
 
Ahh, apparently TeXing does make it clearer. I guess I didn't read your first answer correctly.

I had forgotten about that Log rule, but I'm still a bit rusty. Do all exponents in a logarithm always get multiplied at the front? It seems that they do.

I appreciate your help.
 
Thanks Dick and SammyS. That is exactly what I was looking for.
 
MacLaddy said:
I had forgotten about that Log rule, but I'm still a bit rusty. Do all exponents in a logarithm always get multiplied at the front? It seems that they do.

Yes, in general

[tex]\log(a^b)=b\cdot\log(a), a>0[/tex]

If we look at some special cases such as b=0, this gives us [tex]\log(a^0)=0\cdot\log(a)=0[/tex] which is true because of the log of 1 is always 0.

b=-1 gives

[tex]\log(a^{-1})=-\log(a)[/tex] which can also be seen in another way as
[tex]\log(a^{-1})=\log(\frac{1}{a})=\log(1)-\log(a)=0-\log(a)[/tex]
 
Thanks Mentallic. I think I'll print that for my notes.