Can somebody teach me about logarithms?

**A Beautiful Mind** · July 10th 2009, 08:28 AM

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.

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

The answer anyway...

**Prove It** · July 10th 2009, 08:43 AM

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

The answer anyway...

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

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

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

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

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

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

**A Beautiful Mind** · July 10th 2009, 08:52 AM

I fixed the third problem.

**e^(i*pi)** · July 10th 2009, 08:54 AM

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

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

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

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

**Prove It** · July 10th 2009, 08:59 AM

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...

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