00:01
In this question, we will be using these two properties of logarithms.
00:06
The logarithm of a times b is the logarithm of a plus the logarithm of b, and the logarithm of a raise to the exponent x is x times the logarithm of a.
00:21
For the first expression, we have logarithm sine theta minus the logarithm of sine theta over 5.
00:29
Remember, we can use our properties to split up this product of two numbers.
00:37
Into two logarithms, and it is a product because the 5 in the denominator is the same thing as 5 to the negative 1 in the numerator.
00:51
Lon of sine theta over 5, we write as long of sine theta times 5 to the negative 1, which is lawn of sine theta minus lawn of 5 to the negative 1, which in turn is lon 5, because of the property involving exponents in the antelaborative.
01:11
And because of that, we find that in the answer, lawn of sign cancels out, and the only number we have left is lon 5.
01:25
Positive lon 5.
01:29
For our next expression, we have lon of 3x squared minus 9x plus lon of 1 over 3x.
01:38
To start off, we realize that we can factor the expression in the first logarithm, this first antilogorism, and we can write this antelogorithm as an exponent to the negative 1, a power with an exponent of negative 1.
01:56
When we factor this antilogram, we find that it equals this expression, and 3x to the negative 1 is the other antelogorithm...