# Can somebody teach me about logarithms?

• Jul 10th 2009, 08:28 AM
A Beautiful Mind
Can somebody teach me about logarithms?
Okay, so I'm just trying to run through this study guide and at the present time I'm on logarithms before it progresses to complex numbers. I just looked at this math latex site I usually go on to try and express the stuff so people know, but I think you'll be able to figure it out. Also, sorry if there any typos in my English...I've got blisters all over my hands and it's even hard to grip a pencil.

Here are some problems I'd really like explained to me:

It says to express as a single logarithm.

$1/2logb(x)-logb(y)-logb(z)
$

$1/3logb(x)-2/3logb(y)$

$2log10(3)+4log10(y)-6log10(z)-8log10(t)$

And then these...

$loga\frac{x^2y}{\sqrt[3]{z}
}$

$log2 \sqrt{ \frac{a^6b^4}{z}
}$

Suppose to express those in terms of logarithms.

I mean, I think I might be able to do the first two out of three...I think it's just rearranging...

01. $1/2logb(x)-logb(y)-logb(z)
$

$logb(x)^1/2-logb(y)-logb(z)=logb( \sqrt{x}/yz)$

02. $1/3logb(x)-2/3logb(y)$

$logx^1/3-logx^1/3=logb[\frac{\sqrt[3]{x}{ \sqrt[3]{y}^2]$

Must somehow have an error in this one.

Quote:

logx^1/3-logx^1/3=logb[\frac{\sqrt[3]{x}{ \sqrt[3]{y}^2
• Jul 10th 2009, 08:43 AM
Prove It
Quote:

Originally Posted by A Beautiful Mind
Okay, so I'm just trying to run through this study guide and at the present time I'm on logarithms before it progresses to complex numbers. I just looked at this math latex site I usually go on to try and express the stuff so people know, but I think you'll be able to figure it out. Also, sorry if there any typos in my English...I've got blisters all over my hands and it's even hard to grip a pencil.

Here are some problems I'd really like explained to me:

It says to express as a single logarithm.

$1/2logb(x)-logb(y)-logb(z)
$

$1/3logb(x)-2/3logb(y)$

$2log(10)+4log10(y)-6log10(z)-8log10(t)$

And then these...

$loga\frac{x^2y}{\sqrt[3]{z}
}$

$log2 \sqrt{ \frac{a^6b^4}{z}
}$

Suppose to express those in terms of logarithms.

I mean, I think I might be able to do the first two out of three...I think it's just rearranging...

01. $1/2logb(x)-logb(y)-logb(z)
$

$logb(x)^1/2-logb(y)-logb(z)=logb( \sqrt{x}/yz)$

02. $1/3logb(x)-2/3logb(y)$

$logx^1/3-logx^1/3=logb[\frac{\sqrt[3]{x}{ \sqrt[3]{y}^2]$

Must somehow have an error in this one while trying t

Hopefully you know the following logarithm rules...

$\log_{a}{a^p} = p$

$a^{\log_{a}{p}} = p$

$\log{1} = 0$

$\log{a} + \log{b} = \log{ab}$

$\log{a} - \log{b} = \log{\frac{a}{b}}$

$\log{a^p} = p\log{a}$

Using these rules we can simplify your logarithmic expressions...

1. $\frac{1}{2}\log_b{x} - \log_b{y} - \log_b{z}$

$= \log_b{x^{\frac{1}{2}}} - (\log_b{y} + \log_b{z})$

$= \log_b{x^{\frac{1}{2}}} - \log_b{yz}$

$= \log_b{\frac{x^{\frac{1}{2}}}{yz}}$.

2. $\frac{1}{3}\log_b{x} - \frac{2}{3}\log_b{y}$

$= \log_b{x^{\frac{1}{3}}} - \log_b{y^{\frac{2}{3}}}$

$= \log_b{\frac{x^{\frac{1}{3}}}{y^{\frac{2}{3}}}}$

$= \log_b{\left(\frac{x}{y^2}\right)^{\frac{1}{3}}}$.

3. I think you've missed some information...

4. $\log_a{\frac{x^2 y}{\sqrt[3]{z}}}$

$= \log_a{\frac{x^2 y}{z^{\frac{1}{3}}}}$

$= \log_a{x^2 y} - \log_a{z^{\frac{1}{3}}}$

$= \log_a{x^2} + \log_a{y} - \frac{1}{3}\log_a{z}$

$= 2\log_a{x} + \log_a{y} - \frac{1}{3}\log_a{z}$.

5. $\log_2{\sqrt{\frac{a^6 b^4}{z}}}$

$= \log_2{\left(\frac{a^6 b^4}{z}\right)^{\frac{1}{2}}}$

$= \log_2{\frac{a^3 b^2}{z^{\frac{1}{2}}}}$

$= \log_2{a^3 b^2} - \log_2{z^{\frac{1}{2}}}$

$= \log_2{a^3} + \log_2{b^2} - \frac{1}{2}\log_2{z}$

$= 3\log_2{a} + 2\log_2{b} - \frac{1}{2}\log_2{z}$.
• Jul 10th 2009, 08:52 AM
A Beautiful Mind
I fixed the third problem.
• Jul 10th 2009, 08:54 AM
e^(i*pi)
Quote:

Originally Posted by Prove It
Hopefully you know the following logarithm rules...

$\log_{a}{a^p} = p$

$a^{\log_{a}{p}} = p$

$\log{1} = 0$

$\log{a} + \log{b} = \log{ab}$

$\log{a} - \log{b} = \log{\frac{a}{b}}$

$\log{a^p} = p\log{a}$

Using these rules we can simplify your logarithmic expressions...

1. $\frac{1}{2}\log_b{x} - \log_b{y} - \log_b{z}$

$= \log_b{x^{\frac{1}{2}}} - (\log_b{y} + \log_b{z})$

e^(i*pi): How did you get to $log_b{y} + log_b{z} = log_b{\frac{y}{z}}$? Surely it is equal to $log_b{yz}$

$= \log_b{x^{\frac{1}{2}}} - \log_b{\frac{y}{z}}$

$= \log_b{\frac{x^{\frac{1}{2}}y}{z}}$.

For number 1 I get
$log_b{x^{\frac{1}{2}}} - log_b{y} - log_b{z}$

$= log_b{x^{\frac{1}{2}}} - (log_b{y} + log_b{z})$

$= log_b{x^{\frac{1}{2}}} - log_b{yz}$

$= log_b\frac{x^{\frac{1}{2}}}{yz}$
• Jul 10th 2009, 08:59 AM
Prove It
Quote:

Originally Posted by e^(i*pi)
For number 1 I get
$log_b{x^{\frac{1}{2}}} - log_b{y} - log_b{z}$

$= log_b{x^{\frac{1}{2}}} - (log_b{y} + log_b{z})$

$= log_b{x^{\frac{1}{2}}} - log_b{yz}$

$= log_b\frac{x^{\frac{1}{2}}}{yz}$

Yes you are right...

Goes to edit post...