### Scribe

Today in class, we learned a new way to represent an exponent. Normally when we look at a power it looks something like this, 32 = 9. Instead of the exponent being 2 it becomes log3 9. So when you look at the power, it looks like this, 3log39 = 9. But Mr. K gave us a power that we could not ascertain the value of the exponent.

10^log (5 + X2) = 5 + X2

From this question, we learned that an exponent is a log.

We also learned how to solve inequality questions. First you solve for x by letting the inequality sign (<, >) just seem like an equals sign. We had a problem that represents that type of question.

Log2 (3X-1) < 4

To solve this problem we let the < sign be an =.

Log2 (3X-1) = 4

24 = 3X-1

16= 3x-1

17=3X

X=17/3 or 5.67

Once you find when x equals 4 you bring the inequality sign back into the equation.

Log2 (3X-1) < 4

Since the log is less than 4, your answer becomes X < 4. But because we are using a log, the value of x cannot make the power equal to or less than 0. So now you have to solve (3X-1) for zero.

3X-1 = 0

3X = 1

X= 1/3

Now that you have the minimum and maximum values for the domain of x you right the final solution:

17/3 >x> 1/3

There was also another problem in class that was scary to look at and it involved the natural log, ln x:

5 + ln x = 14/ln x

To solve this you first have to multiply both sides by ln x

ln x(5 + ln x) = (14/ln x)ln x

5 ln x + (ln x)2 = 14

When you multiply ln x by ln x it becomes (ln x)2 not 2 ln x.

5 ln x + (ln x)2 = 14

(ln x)2 + 5 ln x – 14 = 0

let ln x = a

a2 + 5a – 14 = 0

(a+7)(a-2) = 0

a= -7, 2

ln x =-7, 2

Sorry about the subscripts and exponents not being the proper size.

Now tomorrow's scribe is ... Sarah =)

10^log (5 + X2) = 5 + X2

From this question, we learned that an exponent is a log.

We also learned how to solve inequality questions. First you solve for x by letting the inequality sign (<, >) just seem like an equals sign. We had a problem that represents that type of question.

Log2 (3X-1) < 4

To solve this problem we let the < sign be an =.

Log2 (3X-1) = 4

24 = 3X-1

16= 3x-1

17=3X

X=17/3 or 5.67

Once you find when x equals 4 you bring the inequality sign back into the equation.

Log2 (3X-1) < 4

Since the log is less than 4, your answer becomes X < 4. But because we are using a log, the value of x cannot make the power equal to or less than 0. So now you have to solve (3X-1) for zero.

3X-1 = 0

3X = 1

X= 1/3

Now that you have the minimum and maximum values for the domain of x you right the final solution:

17/3 >x> 1/3

There was also another problem in class that was scary to look at and it involved the natural log, ln x:

5 + ln x = 14/ln x

To solve this you first have to multiply both sides by ln x

ln x(5 + ln x) = (14/ln x)ln x

5 ln x + (ln x)2 = 14

When you multiply ln x by ln x it becomes (ln x)2 not 2 ln x.

5 ln x + (ln x)2 = 14

(ln x)2 + 5 ln x – 14 = 0

let ln x = a

a2 + 5a – 14 = 0

(a+7)(a-2) = 0

a= -7, 2

ln x =-7, 2

Sorry about the subscripts and exponents not being the proper size.

Now tomorrow's scribe is ... Sarah =)

## 2 Comments:

At 7:43 PM CDT, Mr. Kuropatwa said…

A great Scribe post Steven!

You've raised the bar another notch!

If you want to write superscipts you have two choices:

(1) 3 squared is 3^2, or

(2) 3<sup>2</sup> will actually make a superscript.

Subscripts can be written like this: log<sub>2</sub>(8)

Try it out. ;-)

Again, way to go on your post. You've set a new standard!

At 9:17 PM CDT, SarahS said…

whoah didn't expect me being scribe tomorrow. it's all good. Great scribing steven! It's like I know exactly what i missed in class if i happened to miss class. But for sure i wouldn't be missing class.

wow people have been setting new standards to follow. I'll try and get my scribe done at least before midnight. But for sure it wouldn't be done earlier done 9 since i have work till then.

Great Job once again.

till my scribe, baboo for now. =D

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