(2 points) 14. Use the properties of logarithms to expand...

(2 points) 14. Use the properties of logarithms to expand the expression as a sum, difference, and/or constant multiple of logarithms: $\log(\frac{yz^2}{\sqrt{x}})$ (6 points) 15. Solve the following for x: a) $2^{(7x+4)} = 64$ b) $e^{2x} - 2e^x - 3 = 0$ c) $\log_3(x+6) + \log_3(x+4) = 1$

Transcript

00:01 The first thing you would want to do with 4a is rewrite this in log form.

00:06 So 2 to the 5th power equals 4x, which 2 to the 5th power is equal to 32.

00:12 So when you divide that 4 over, you'll end up with x equals 8.

00:19 You can do the same thing on letter b.

00:22 It's all the same work, except it's x to the 6th power equals 64, which is not as easy.

00:30 I think what i would do is just tell students to change 64 to be a base of 2, 2 to the 6th power.

00:36 Or no, that's probably not the best advice.

00:38 Let me cross that out.

00:39 Better advice would be doing the 6th root or the 1 6th power.

00:45 And you can actually get two answers here because it's an even exponent.

00:49 It could be positive or negative 2 because negative 2 to the 6th power would give you positive 64.

00:55 And then letter c is just 2 to the 7th power is equal to x.

00:58 And i'm pretty sure 2 to the 7th power, you type that in, is 128.

01:05 So that's all you have to do on that one.

01:08 Now number 5 is a little bit trickier.

01:12 Looks like they want you to simplify, i mean guess that means they want you to condense it down, and since all of them are subtraction, you can combine subtraction together using division, and what you can actually do on this first one is cancel out an x.

01:27 So you're looking at log base 2 of y over x.

01:32 And that's it...

Question

(2 points) 14. Use the properties of logarithms to expand the expression as a sum, difference, and/or constant multiple of logarithms: $\log(\frac{yz^2}{\sqrt{x}})$ (6 points) 15. Solve the following for x: a) $2^{(7x+4)} = 64$ b) $e^{2x} - 2e^x - 3 = 0$ c) $\log_3(x+6) + \log_3(x+4) = 1$

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