Logarithm Questions - Solve Now!

  • Thread starter Thread starter Jouster
  • Start date Start date
  • Tags Tags
    Logarithm
AI Thread Summary
To simplify the logarithmic equation, apply the fundamental principle of logarithms, which states that log_a(bc) = log_a(b) + log_a(c). The equation can be rewritten as log_{3a}9 = (log_a9)/(log_a3a) = (log_a3^2)/(log_aa + log_a3) = (2log_a3)/(1 + log_a3). It's important to consistently use LaTeX for mathematical expressions instead of images for clarity. Engaging with the logarithmic principles will enhance understanding and problem-solving skills. Mastering these concepts is crucial for effective algebraic manipulation.
Jouster
Messages
5
Reaction score
0
Homework Statement
By changing the base of log[SUB]3a[/SUB]9, express (log3a9)(1+loga3) as a single logarithm to base a. I don't know what to do to simplify the equation further
Relevant Equations
Logarithms
1671093569418.png
 
Last edited:
Physics news on Phys.org
Jouster said:
I don't know what to do to simplify the equation further
Minimize your algebra by applying what you know about logarithms to finish rewriting your first line:
$$\log_{3a}9=\frac{\log_{a}9}{\log_{a}3a}=\frac{\log_{a}3^{2}}{\log_{a}a+\log_{a}3}=\frac{2\log_{a}3}{1+\log_{a}3}$$
 
@Jouster: this is the latest of several threads you've started about logarithms where you have not applied the fundamental principle of logarithms: \log_a(bc) = \log_a(b) + \log_a(c). You need to get into the habit of applying this principle before asking for help.
 
Also @Jouster -- Please post your math using LaTeX, not in images. See the "LaTeX Guide" link below the Edit window for more information. Thank you.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top