SOLUTION: show that {{2log(16/15)+log(25/24)-log(32/27)}}=0

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Question 595609: show that
{{2log(16/15)+log(25/24)-log(32/27)}}=0
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
2log%28%2816%2F15%29%29%2Blog%28%2825%2F24%29%29-log%28%2832%2F27%29%29=0
To find out if the left side of the equation is really equal to zero we need to combine/condense them down to a single logarithm. And since the arguments of the logarithms are different, we will need to use the following properties of logs:
  • log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29
  • log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29

These properties require that the coefficients of the logs be 1's. Fortunately there is another property of logarithms, q%2Alog%28a%2C+%28p%29%29+=+log%28a%2C+%28p%5Eq%29%29, which allows us to "move" a coefficient into the argument as its exponent.

So we will start by using the third property to move the 2 in front of the first log:
log%28%28%2816%2F15%29%5E2%29%29%2Blog%28%2825%2F24%29%29-log%28%2832%2F27%29%29=0

Now we can start combining terms. The first two logs have a "+" between them so we will use the first property (which also has a "+" between the logs):
log%28%28%2816%2F15%29%5E2%2A%2825%2F24%29%29%29-log%28%2832%2F27%29%29=0

The remaining logs have a "-" between them so we will use the second property (which also has a "-" between the logs):
log%28%28%2816%2F15%29%5E2%2A%2825%2F24%29%2F%2832%2F27%29%29%29=0

Now we will set about to simplify the big fraction inside the log. Since dividing is the same as multiplying by a reciprocal, I'm going to rewrite the "divide by 32/27" into "multiply by 27/32":
log%28%28%2816%2F15%29%5E2%2A%2825%2F24%29%2A%2827%2F32%29%29%29=0
Next I'm going to start canceling factors that are common to the numerators and denominators. In order to see all the factors I am going to first rewrite the squared fraction without an exponent:
log%28%28%2816%2F15%29%2A%2816%2F15%29%2A%2825%2F24%29%2A%2827%2F32%29%29%29=0
and then I'll factor the numerators and denominators:

Now we can start canceling:

As you can see, everything cancels out! So all we are left with is:
log%28%281%29%29+=+0
And since the zero power of 10 is 1, log(1) is 0:
0 = 0 Check!

Note: The bases of the logarithms could have been any number (as long as all three bases were the same) since the zero power of any base is equal to 1!