SOLUTION: using the law of logarithms how would you rewrite the following equation as one logarithm? {{{ln5+ln5^2+ln5^3-ln5^6}}} I keep getting 0 but the answers in the back of the boo

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: using the law of logarithms how would you rewrite the following equation as one logarithm? {{{ln5+ln5^2+ln5^3-ln5^6}}} I keep getting 0 but the answers in the back of the boo      Log On


   

Question 231883: using the law of logarithms how would you rewrite the following equation as one logarithm?
ln5%2Bln5%5E2%2Bln5%5E3-ln5%5E6
I keep getting 0 but the answers in the back of the book is lin1+=+0 is this equivilant to 0? How did they get this answer?
(those are natural log functions, I thought it would much be harder to understand if written using log form)

Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
Here's a couple of ways to simplify this expression:
  • Use the following property of logarithms, log%28a%2C+%28x%5Ey%29%29+=+y%2Alog%28a%2C+%28x%29%29, to "move" the exponent in the argument out in front, which changes
    ln%285%29%2Bln%285%5E2%29%2Bln%285%5E3%29-ln%285%5E6%29
    into
    ln%285%29+%2B+2ln%285%29+%2B+3ln%285%29+-+6ln%285%29
    Adding like terms:
    %281+%2B+2+%2B+3+-+6%29ln%285%29
    0%2Aln%285%29
    which simplifies to:
    0
  • Use the following properties of logarithms:
    • log%28a%2C+%28x%29%29+%2B+log%28a%2C+%28y%29%29+=+log%28a%2C+%28x%2Ay%29%29
    • log%28a%2C+%28x%29%29+-+log%28a%2C+%28y%29%29+=+log%28a%2C+%28x%2Fy%29%29
    to "combine" the logarithms. This will change
    ln%285%29%2Bln%285%5E2%29%2Bln%285%5E3%29-ln%285%5E6%29
    into
    ln%28%285%2A5%5E2%2A5%5E3%29%2F5%5E6%29
    Using our rules for exponents to multiply the numerator:
    ln%285%5E6%2F5%5E6%29
    And since any non-zero number divided by itself is 1:
    ln%281%29
    This expression represents the exponent for e which results in 1. Since any non-zero number, including e, to the zero power is 1, 0+=+ln%281%29+=+log%28%281%29%29+=+log%282%2C+%281%29%29+=+log%283%2C+%281%29%29 = ..., we get
    0