SOLUTION: use the properties of logarithms to write the single logarithm as a sum and/or difference of logarithms with no exponents in the simplified expression: LOG((2x^2-3x+1)/(e^x(5x-1

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: use the properties of logarithms to write the single logarithm as a sum and/or difference of logarithms with no exponents in the simplified expression: LOG((2x^2-3x+1)/(e^x(5x-1      Log On


   

Question 261784: use the properties of logarithms to write the single logarithm as a sum and/or difference of logarithms with no exponents in the simplified expression:
LOG((2x^2-3x+1)/(e^x(5x-1)^2))
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%28%28%282x%5E2-3x%2B1%29%2F%28%28e%5Ex%29%285x-1%29%5E2%29%29%29

There are three basic properties of logarithms which can be use to take logarithms of one form and rewrite them as equivalent logarithms in a different form:
  • log%28a%2C+%28p%2Aq%29%29+=+log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29
  • log%28a%2C+%28p%2Fq%29%29+=+log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29
  • log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29

Since the argument of your logarithm is one big fraction, we can use the middle property above to rewrite the simgle logarithm as a difference of logarithms:
log%28%282x%5E2-3x%2B1%29%29+-+log%28%28%28e%5Ex%29%285x-1%29%5E2%29%29
The second logarithm is the log of a product. The first property above can be used to split that into the sum to two logs. (Note: Because of the minus sign in front of the second log, I will use parentheses when I make my substitution. It is generally a good idea to use parentheses when making substitutions, especially when more than one term is involved.)
log%28%282x%5E2-3x%2B1%29%29+-+%28log%28%28e%5Ex%29%29+%2B+log%28%28%285x-1%29%5E2%29%29%29
which simplifies to:
log%28%282x%5E2-3x%2B1%29%29+-+log%28%28e%5Ex%29%29+-+log%28%28%285x-1%29%5E2%29%29%29
The arguments of the second and third logarithms have exponents. We can use the third property above to move the exponents out in front:
log%28%282x%5E2-3x%2B1%29%29+-+x%2Alog%28%28e%29%29+-+2log%28%285x-1%29%29
The only thing left is the first logarithm. It has an exponent. But its exponent is not on the entire argument. It is only on the first term of the argument. So we cannot use the third property on it like we just finished doing with the second and third logs. The key to rewriting this log is to recognize that
  • The first property above allows us to split the log of a product.
  • The argument in the first log can be factored. IOW: we can rewrite it as a product.
  • Exponents of factors are lower than the exponents in the unfactored expression.

Put all these together and we have a solution to the exponent in the argument (which we do not want). So we factor the argument:
log%28%28%282x-1%29%28x-1%29%29%29+-+x%2Alog%28%28e%29%29+-+2log%28%285x-1%29%29
(See no (visible) exponents!) Now we can use the first property to rewrite the first log as a sum:
log%28%282x-1%29%29+%2B+log%28%28x-1%29%29+-+x%2Alog%28%28e%29%29+-+2log%28%285x-1%29%29