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

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- June 7th 2014, 04:45 PMLaneRendellCondesing 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

- June 7th 2014, 05:06 PMPlatoRe: Condesing this Logarithmic Expression
- June 7th 2014, 05:12 PMLaneRendellRe: Condesing this Logarithmic Expression
Yep, golden. Thank you!

- June 7th 2014, 05:50 PMHallsofIvyRe: Condesing this Logarithmic Expression
So what

**can**you do with that? You know the "laws of logarithms", and don't you? - June 7th 2014, 06:00 PMLaneRendellRe: Condesing this Logarithmic Expression
Right my first thought is to bring the and over from in front of the logs,

which would be +

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

Which simplifies to:

Edit: So, becomes

As for the first term I'm confused on how to tackle that. - June 7th 2014, 07:21 PMSorobanRe: Condesing this Logarithmic Expression
Hello, LaneRendell!

Okay, you gave it a try.

I'll do it baby steps . . .

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- June 7th 2014, 07:35 PMJeffMRe: 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? - June 7th 2014, 07:42 PMLaneRendellRe: Condesing this Logarithmic Expression
- June 7th 2014, 07:43 PMLaneRendellRe: 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.

- June 7th 2014, 08:08 PMJeffMRe: 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. - June 7th 2014, 09:03 PMLaneRendellRe: Condesing this Logarithmic Expression
Got it to work using Soroban's explanation. My confusion was the step:

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! :) - June 7th 2014, 10:16 PMProve ItRe: 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. - June 16th 2014, 02:31 AMmuhsinRe: Condesing this Logarithmic Expression
well done...

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