# Log Simplification Function

1. Jan 18, 2012

1. The problem statement, all variables and given/known data

Assume $log_bx=0.36, log_by=0.56, log_bz=0.83$

Evaluate the following expressions

$$log_b\frac{\sqrt{xy}}{z}$$

2. Relevant equations

3. The attempt at a solution

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

However, I have no idea what to do with the $\sqrt{x}$ 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

2. Jan 18, 2012

### Dick

log(sqrt(x))=log(x^(1/2))=(1/2)*log(x). Does that help?

3. Jan 18, 2012

I guess I should clarify what I am asking.

If $log_bx = 0.36,$ then does $log_b\sqrt{x}=\sqrt{0.36}$

4. Jan 18, 2012

### Dick

Nooo. $log_b\sqrt{x}=log_b x^\frac{1}{2}=\frac{1}{2}*log_b{x}$. Does TeXing it make it clearer?

5. Jan 18, 2012

### SammyS

Staff Emeritus

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

For one thing, $\sqrt{x}=x^{1/2}\,.$

6. Jan 18, 2012

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.

7. Jan 18, 2012

Thanks Dick and SammyS. That is exactly what I was looking for.

8. Jan 18, 2012

### Mentallic

Yes, in general

$$\log(a^b)=b\cdot\log(a), a>0$$

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

b=-1 gives

$$\log(a^{-1})=-\log(a)$$ which can also be seen in another way as
$$\log(a^{-1})=\log(\frac{1}{a})=\log(1)-\log(a)=0-\log(a)$$

9. Jan 18, 2012