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
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.
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?
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.
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!
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.
well done...
Soran University