SOLUTION: Using the properties of logarithms which single term does not contain a logarithm: e^ln(3x^5). (1)3x^5, (2)x^15 (3) x^8 or (4) x^5/3?

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Using the properties of logarithms which single term does not contain a logarithm: e^ln(3x^5). (1)3x^5, (2)x^15 (3) x^8 or (4) x^5/3?      Log On


   

Question 202905: Using the properties of logarithms which single term does not contain a logarithm: e^ln(3x^5). (1)3x^5, (2)x^15 (3) x^8 or (4) x^5/3?
Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
e%5Eln%283x%5E5%29
Here are two solutions:
  1. If you really understand what logarithms represent then the answer is obvious. In general log%28a%2C+%28x%29%29 represents "the exponent one could put on 'a' to get 'x'. In your problem you have ln%283x%5E5%29 which represents "the exponent one could put on 'e' to get "3x%5E5". And where do we find ln%283x%5E5%29? As an exponent on 'e'! So e%5Eln%283x%5E5%29 is, by the very definition and meaning of logarithms: 3x%5E5!
  2. Let's say the y+=+e%5Eln%283x%5E5%29. Now let's find the ln of both sides:
    ln%28y%29+=+ln%28e%5E%28ln%283x%5E5%29%29%29
    Using the ln%28a%5Eb%29+=+b%2Aln%28a%29 property we can rewrite the right side:
    ln%28y%29+=+ln%283x%5E5%29%2Aln%28e%29
    The ln(e) is 1 so now we have
    ln%28y%29+=+ln%283x%5E5%29
    Since the ln's are equal, the arguments must be equal:
    y+=+3x%5E5
    Substituting back in for y using y+=+e%5Eln%283x%5E5%29 we get:
    e%5Eln%283x%5E5%29+=+3x%5E5