SOLUTION: Express in terms of logarithms without exponents. log_c(4 square root m^12 n^8, which m and n is over (or divided by) c^7 What is the equivalent sum or difference of logari

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Express in terms of logarithms without exponents. log_c(4 square root m^12 n^8, which m and n is over (or divided by) c^7 What is the equivalent sum or difference of logari      Log On


   

Question 423872: Express in terms of logarithms without exponents.
log_c(4 square root m^12 n^8, which m and n is over (or divided by) c^7
What is the equivalent sum or difference of logarithms?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
"log_c(4 square root m^12 n^8, which m and n is over (or divided by) c^7" is a very confusing expression. There are literally dozens of things it could mean.

Please re-post your question. Here are some tips for when you do:
  • Get some help and learn how to read these expressions so you can describe them, partially or completely, in plain English.
  • Logarithms are especially hard to type. I recommend
    • Partial or full English descriptions. For example log%284%2C+%28x%5E2%2B4x-3%29%29 could be described as "the base 4 log of (x^2+4x-3)" and/or...
    • Learning Algebra.com's syntax for displaying logarithms. Click on the "Show source" link which you will find just above this answer. You will then see what I typed to get Algebra.com to display the logarithm above so nicely. You can use this syntax when you post your problems.
  • Use parentheses around...
    • Numerators
    • Denominators
    • Radicands (The expression inside a radical is called a radicand.)
    • Function arguments. (In log%284%2C+%28x%5E2%2B4x-3%29%29 the function is log and its argument is x^2+4x-3.)
    • Multiple term exponents
    When in doubt, use parentheses to group parts of an expression that should be treated as a unit.

The bottom line: The more clearly understandable the problem is, the faster you will get a response from one of the tutors.