00:02
Today i'm going to practice simplifying some logarithmic expressions using the properties of logs.
00:08
So for example, if we have the natural log of sine of theta minus the natural log of sine of theta over five, we can then actually figure out what this simplifies to.
00:28
So because we have a subtraction sign, we know we can rewrite this as the natural log of sine.
00:37
Here being our numerator over this whole thing here as our denominator, sine of theta over five.
00:53
And when you get a fraction over a fraction, we have this main fraction here and another fraction down here.
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We can flip the bottom fraction and multiply on the inside here.
01:04
So we get sine of theta times five over sine of theta, and therefore the signs cancel.
01:13
So the natural log of 5 would be our solution.
01:18
And this, of course, assumes that sine theta is on the appropriate domain, that it's greater than 0.
01:24
So, let's look at another example.
01:26
Let's say we do the natural log of, how about 3x squared minus 9x plus the natural log of 1 over 3x.
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Because we have a plus sign here, we can multiply the quantities inside the natural logs.
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So we get 3x squared minus 9x times 1 over 3x.
01:57
And so we can simplify that and say the natural log of 3x squared over 3x minus 9x over 3x.
02:09
Simplifying what's inside 3x squared divide by 3x, 3 is cancel, the x is canceled down 1, leaving just x...