### scribe, logs and exponents.

I'm the scribe today, so let's start (:

For the morning period - Mr. K was handing back all our quizzes, pre-tests, tests and so forths back to us. And he told us our mark currently in precal. Mr. K said that the number doesn't prove to anyone that it's how smart you are at math. The mark represents how much time and effort you put into this course. Yes, and then we started going over the tests which we had questions about..

For the afternoon class - Mr. K put questions on the board for us to do, and here they are:

Find the: **A.) root B.) inverse function**

i.) y=8^{x-2} ii.) y+1=log_{8}(x-2)

i.) y=8^{x-2}**A.)** no roots, because it doesn't hit/cross the x-axis.**B.)** x=8^{y-2} ; switch the x and the y variable.

logx=log_{8}^{(y-2)} ; log both sides.

logx=(y-2)log_{8} ; bring down the (y-2)

logx/log_{8}=y-2 ; divide both sides by log_{8}

logx/log_{8}+2=y ; bring the 2 to the left side, which becomes positive.

log_{8}x+2=y

As you noticed, in the original question, it was minus two and for the inverse, it became positive two.

ii.) y+1=log_{8}(x-2)**A.)** 1=log_{3}(x-2) ; y = 0

3^{1}=x-2 ; change to exponential.

5=x**B.)** x+1=log_{3}(y-2) ; switch the x and the y variable.

3^{(x+1)}=y-2 ; change to exponential.

3^{(x+1)}+2=y

Same for this one, in the original question, it was a minus two and for the inverse you get a positive two.

**LOGARITHM IS AN EXPONENT!!**Solve for x: A.) log

_{5}(2x-1)+log

_{5}(x-2)=1

B.) log

_{2}(2-2x)+log

_{2}(1-x)=5 C.) 2

^{x}=7

A.) log

_{5}(2x-1)+log

_{5}(x-2)=1

log

_{5}(2x-1)(x-2)=1

(2x-1)(x-2)=5

^{1}

2x

^{2}-5x+2=5

2x

^{2}-5x-3=0

(2x+1)(x-3)=0

2x+1=0 x-3=0

You have to plug the numbers into the original equation to find out which one you can reject. So, if you plug in -1/2 into the equation.. (2(-1/2)-1), you would get -2 and (-1/2-2), you would get -2 1/2. SO that means it won't work, so it gets rejected. And if you plug in 3 into the equation.. (2(3)-1), you would get 5 and (3-2), you would get one. WHICH means, it will work (:

B.) log

_{2}(2-2x)+log

_{2}(1-x)=5

log

_{2}(2-2x)(1-x)=5

(2-2x)(1-x)=2

^{5}

2-4x+2x

^{2}=32

2x

^{2}-4x-30=0

(2x+6)(x-5)=0

2x+6=0 x-5=0

x=-3

Same for this one, plug in -3 into the equation.. (2-2(-3)), which you would get eight and (1-(-3)), which you would get four. SO that means, it will work. And if you plug in 5 into the equation.. (2-2(5)), which would equal to -8 and (1-5), which would equal to -4. WHICH mean, it doesn't work out, so it gets rejected.

**CHECK**before you REJECT (:

C.) 2

^{x}=7 ; log both sides.

log2

^{x}=log7 ; bring down the x variable.

xlog2=log7 ; divide by log2 so you could isolate x

x=log7/log2

log

_{2}7=x

Then Mr. K talked about the compound interest formula.

Which is A=P(1+r/n)

^{nt}.

OH YEAH, A= accumulated amount of money

P= principal amount of money

r= rate in decimal form (interest rate)

n= number of compounding periods in a year

t= time in years

Mr. K then said he'll offer us a deal, if we invest one dollar and get 100% interest.

(1+(1/1))

^{1}=2

(1+(1/2))

^{2}=2.25

(1+(1/100))

^{100}=2.704813829

(1+(1/1000000))

^{1000000}=2.718280469

(1+(1/1000000000))

^{1000000000}=2.718281827

And so on, and so on..

The offer may sound good, but once you calculate it all out, it's really not alot. SO Mr. K talked about " e " (Euler's number), which equals to 2.718281828149

" e " can be written as an exponent as well. Which would be y=e

^{x}and as a log. Which is log

_{e}x=y ..but we don't write it that way. SO it's written like this.. lnx=y which is a natural log of x.

Yeah, that's about it.. Um, the scribe for tomorrow is.. tennyson.

gooday&goodnight.

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