1. ## 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.

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

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

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

And then these...

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

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

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

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

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

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

Must somehow have an error in this one.

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

2. 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.

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

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

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

And then these...

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

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

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

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

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

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

Must somehow have an error in this one while trying t

Hopefully you know the following logarithm rules...

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

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

$\displaystyle \log{1} = 0$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

3. I fixed the third problem.

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

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

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

$\displaystyle \log{1} = 0$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle = log_b\frac{x^{\frac{1}{2}}}{yz}$
Yes you are right...

Goes to edit post...