Log Simplification Strategies for Evaluating Expressions with Square Roots

In summary, the conversation discusses the evaluation of the expression log_b\frac{\sqrt{xy}}{z} using the logarithm rule \log(a^b)=b\cdot\log(a). The participants also clarify the use of exponents in logarithms and provide examples to solidify the concept.
  • #1
MacLaddy
Gold Member
291
11

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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  • #2
log(sqrt(x))=log(x^(1/2))=(1/2)*log(x). Does that help?
 
  • #3
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]
 
  • #4
Nooo. [itex]log_b\sqrt{x}=log_b x^\frac{1}{2}=\frac{1}{2}*log_b{x}[/itex]. Does TeXing it make it clearer?
 
  • #5
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]
 
  • #6
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.
 
  • #7
Thanks Dick and SammyS. That is exactly what I was looking for.
 
  • #8
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]
 
  • #9
Thanks Mentallic. I think I'll print that for my notes.
 

What is a log simplification function?

A log simplification function is a mathematical operation that simplifies logarithmic expressions by combining terms and reducing the number of logarithms present.

Why do we use log simplification functions?

We use log simplification functions to make logarithmic expressions easier to work with and to find simpler solutions for equations involving logarithms.

What are the basic rules for simplifying logarithms?

The basic rules for simplifying logarithms include the power rule, product rule, quotient rule, and change of base rule. These rules allow us to manipulate the expression to combine terms and reduce the number of logarithms present.

What are the common mistakes made when simplifying logarithms?

Some common mistakes when simplifying logarithms include not applying the rules correctly, forgetting to simplify further, or incorrectly combining terms. It is important to be careful and check your work when simplifying logarithms.

How can we use log simplification functions in real life?

Log simplification functions are used in many fields of science and engineering, such as in chemistry to calculate pH levels and in physics to solve exponential decay problems. They are also used in finance and economics for calculating compound interest and in computer science for data compression and storage.

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