# Condesing this Logarithmic Expression

• Jun 7th 2014, 03:45 PM
LaneRendell
Condesing this Logarithmic Expression
Typically these don't give me much trouble but I'm having trouble with the exponents. If someone could give me a very slow walkthrough I'd be very grateful. :)

Problem: Using the logarithmic properties condense the following logarithm

$\displaystyle \frac{-2}{3}\log_5(5m^2)+\frac{1}{2}\log_5(25m^2)$
• Jun 7th 2014, 04:06 PM
Plato
Re: Condesing this Logarithmic Expression
Quote:

Originally Posted by LaneRendell
Typically these don't give me much trouble but I'm having trouble with the exponents. If someone could give me a very slow walkthrough I'd be very grateful. :)

Problem: Using the logarithmic properties condense the following logarithm

$\displaystyle \frac{-2}{3}\log_5(5m^2)+\frac{1}{2}\log_5(25m^2)$

I have tried to cleanup your lousy LaTeX coding.
Is that the that a correct reading?
• Jun 7th 2014, 04:12 PM
LaneRendell
Re: Condesing this Logarithmic Expression
Yep, golden. Thank you!
• Jun 7th 2014, 04:50 PM
HallsofIvy
Re: Condesing this Logarithmic Expression
So what can you do with that? You know the "laws of logarithms", $\displaystyle log(a^b)= b log(a)$ and $\displaystyle log(a)+ log(b)= log(ab)$ don't you?
• Jun 7th 2014, 05:00 PM
LaneRendell
Re: Condesing this Logarithmic Expression
Right my first thought is to bring the $\displaystyle -\frac{2}{3}$ and $\displaystyle \frac{1}{2}$ over from in front of the logs,

which would be $\displaystyle \log_5(5m^2)^(-\frac{2}{3})$+$\displaystyle \log_5(25m^2)^(\frac{1}{2})$

I also recognize that on the second term (as well as the first but it doesn't appear to work out as nicely) you could do $\displaystyle \sqrt\log_5(25m^2)$

Which simplifies to: $\displaystyle \log_5(5m)$

Edit: $\displaystyle \log_a(a^k) = k$ So, $\displaystyle \log_5(5m)$ becomes $\displaystyle 1 + \log_5(m)$

As for the first term I'm confused on how to tackle that.
• Jun 7th 2014, 06:21 PM
Soroban
Re: Condesing this Logarithmic Expression
Hello, LaneRendell!

Okay, you gave it a try.
I'll do it baby steps . . .

Quote:

$\displaystyle -\tfrac{2}{3}\log_5(5m^2)+\tfrac{1}{2}\log_5(25m^2)$

$\displaystyle \log_5(5m^2)^{\text{-}\frac{2}{3}} + \log_5(25m^2)^{\frac{1}{2}}$

. . $\displaystyle =\;\log_5\left[5^{\text{-}\frac{2}{3}}(m^2)^{\text{-}\frac{2}{3}}\right] + \log_5\left[25^{\frac{1}{2}}(m^2)^{\frac{1}{2}}\right]$

. . $\displaystyle =\;\log_5\left(5^{\text{-}\frac{2}{3}}m^{\text{-}\frac{4}{3}}\right) + \log_5(5m)$

. . $\displaystyle =\;\log_5\left(5^{\text{-}\frac{2}{3}}m^{\text{-}\frac{4}{3}}\cdot 5m\right)$

. . $\displaystyle =\;\log_5\left(5^{\frac{1}{3}}m^{\text{-}\frac{1}{3}}\right)$

. . $\displaystyle =\;\log_5\left(5m^{\text{-}1}\right)^{\frac{1}{3}}$

. . $\displaystyle =\;\tfrac{1}{3}\log_5(5m^{\text{-}1})$

. . $\displaystyle =\;\tfrac{1}{3}\left[\log_5(5) + \log_5(m^{\text{-}1})\right]$

. . $\displaystyle =\;\tfrac{1}{3}\left[1 + \log_5(m^{\text{-}1})\right]$

. . $\displaystyle =\;\tfrac{1}{3}\left(1 - \log_5m)$

• Jun 7th 2014, 06:35 PM
JeffM
Re: Condesing this Logarithmic Expression
You can do this multiple ways, but personally I'd start by separating the m terms and the powers of 5 because the logs to the base 5 of powers of 5 are integers, and I can (usually) cope with small integers.

$-\ \dfrac{2}{3} * log_5(5m^2)+ \dfrac{1}{2} * log_5(25m^2) =$

$-\ \dfrac{2}{3}\{log_5(5) + log_5(m^2)\} + \dfrac{1}{2}\{log_5(5^2) + log(m^2)\}$. Now what?
• Jun 7th 2014, 06:42 PM
LaneRendell
Re: Condesing this Logarithmic Expression
Quote:

Originally Posted by Soroban
Hello, LaneRendell!

Okay, you gave it a try.
I'll do it baby steps . . .

$\displaystyle \log_5(5m^2)^{\text{-}\frac{2}{3}} + \log_5(25m^2)^{\frac{1}{2}}$

. . $\displaystyle =\;\log_5\left[5^{\text{-}\frac{2}{3}}(m^2)^{\text{-}\frac{2}{3}}\right] + \log_5\left[25^{\frac{1}{2}}(m^2)^{\frac{1}{2}}\right]$

. . $\displaystyle =\;\log_5\left(5^{\text{-}\frac{2}{3}}m^{\text{-}\frac{4}{3}}\right) + \log_5(5m)$

. . $\displaystyle =\;\log_5\left(5^{\text{-}\frac{2}{3}}m^{\text{-}\frac{4}{3}}\cdot 5m\right)$

Get lost right after where I cut it, not sure how you distributed those exponents
• Jun 7th 2014, 06:43 PM
LaneRendell
Re: Condesing this Logarithmic Expression
JeffM: I'd probably simplify those two log base 5's (the one's that are integers) into the integers 1 and 2.
• Jun 7th 2014, 07:08 PM
JeffM
Re: Condesing this Logarithmic Expression
Given that soroban has worked it out for you

$-\ \dfrac{2}{3}\{log_5(5) + log_5( m^2)\} + \dfrac{1}{2}\{log_5(5^2) + log_5(m^2)\} =$

$-\ \dfrac{2}{3}\{1 + log_5( m^2)\} + \dfrac{1}{2}\{2 + log_5(m^2)\} =$

$1 -\dfrac{2}{3} + log_5(m^2)\left(\dfrac{1}{2} - \dfrac{2}{3}\right) =$

$\dfrac{1}{3} + log_5(m^2) * \left(-\ \dfrac{1}{6}\right) =$

$\dfrac{1}{3} + 2 * log_5(m) * \left(-\ \dfrac{1}{6}\right) =$

$\dfrac{1}{3} - log_5(m) * \dfrac{1}{3} =$

$\dfrac{1}{3} * \{1 - log_5(m)\}.$ Same answer, slightly different (and to me somewhat simpler) method.
• Jun 7th 2014, 08:03 PM
LaneRendell
Re: Condesing this Logarithmic Expression
Got it to work using Soroban's explanation. My confusion was the step: $\displaystyle \log_5(5^-\frac{2}{3}m^-\frac{4}{3}*5m)$

I failed to recognize that the first term and second term had the same bases so you can use the exponent rule where multiplication means adding exponents, and that you add the exponent 3/3 to both the -4/3 and -1/3

Also thank you to both of you for your help! Feels so good to have been steered in the right direction! :)
• Jun 7th 2014, 09:16 PM
Prove It
Re: Condesing this Logarithmic Expression
First of all, in the LaTeX code, if you put your exponent inside curly brackets it will all be made superscript, e.g. you need a power of -2/3 so write ^{-\frac{2}{3}}.

Second, I slightly disagree with Soroban's solution, as \displaystyle \begin{align*} \left( m^2 \right) ^{\frac{1}{2}} = |m| \end{align*}, not m. Since we aren't told anything about m, we are required to assume it's any value in the implied domain of the function. Since logarithms are only defined for the positive numbers, we require the stuff inside it to be positive. But the stuff inside is gotten by squaring a number, which turns any nonzero number positive. Thus m can be any nonzero number.
• Jun 16th 2014, 01:31 AM
muhsin
Re: Condesing this Logarithmic Expression
Quote:

Originally Posted by LaneRendell
Typically these don't give me much trouble but I'm having trouble with the exponents. If someone could give me a very slow walkthrough I'd be very grateful. :)

Problem: Using the logarithmic properties condense the following logarithm

$\displaystyle \frac{-2}{3}\log_5(5m^2)+\frac{1}{2}\log_5(25m^2)$

well done...
Soran University