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

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

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?

Re: Condesing this Logarithmic Expression

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?

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.

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}) $

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

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

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

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?

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

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.

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.

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! :)

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.

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