### blogging logarhythms

Ok sorry for the late blog. First things first using logs remember these

Use this with the quotient questions

Log

_{a}(m/n)=log

_{a}m - log

_{a}n

Use this with the multiplication questions

Log

_{a}(mn)= log

_{a}m+log

_{a}n

This one I'm not sure when to use it but it's tricky

cLog

_{a}m=log

_{a}m

^{a}

express as a logarhythm of a single number or expression

a)log

_{a}2+ log

_{a}3

Log

_{a}(mn)

=log

_{a}(2*3)

b)log

_{a}5-log

_{a}4

Log

_{a}(m/n)

=log

_{a}(5/4)

c)3logLog

_{a}2

log

_{a}m

^{a}

log

_{a}3

^{2}

d)-Log

_{a}(1/6)

log

_{a}m

^{a}

log

_{a}(1/6)

^{-1}

e)1/2log

_{b}m + 1/2 log

_{b}n

Log

_{a}(mn)

There are two ways to do this you can either put it as log

_{b}Ö(mn) or

Log

_{b}m

^{1/2}n

^{1/2}

=log

_{b}mn

^{1/2}

f)1/2(log

_{b}x-log

_{b}4)

Log

_{a}(m/n)

=log

_{b}Ö(x/4)

for these ones let C=Log

_{3}(10) and let d=Log

_{3}(5) and express in terms of C and D which means that when you simplify it it and it looks like either C or D then input it in for example Log

_{3}(10)+Log

_{3}(5) then replace it and you get C+D if you don't get it I'm sorry I don't really explain well but here are the questions you'll get and answers you'll see what I mean...

a)Log

_{3}250

=log

_{3}

^{(52*10)}

=log

_{3}5

^{2}+log

_{3}10

=2log

_{3}5+log

_{3}10

+2D+C

b)log

_{3}log

^{2}

=log

_{3}10/5

=log

_{3}10-log

_{3}5

=C-D

c)log

_{3}Ö(5)

=log

_{3}(5)

^{1/2}

=1/2log

_{3}5

d)log

_{3}1/100

=-2log

_{3}10

=log

_{3}(1/100

^{2})

=log

_{3}1-log

_{3}10

^{2}

it just says to expand so these questions

a)log

_{2}m

^{5}n

^{4}

=5log

_{2}m+4log

_{2}n

b)log

_{2}(mn)

^{3}

=3log

_{2}m+3log

_{2}n

c)log

_{2}

^{3}Ö(m

^{2}n)

2/2log

_{2}m+1/3log

_{2}n

d)log

_{2}(Ö(m/n

^{3}))

=1/2log

_{2}m-3/2log

_{2}n

solve these questions plus mr.K used something called the anti-log as shown on question A it gets rid of the log

a)log

_{a}x=log

_{a}a+log

_{a}5

=anti-log

_{a}(log

_{a}x)=anti-log

_{a}(log

_{a}45)

x=45

b)log

_{a}x=log

_{a}

^{3}+log

_{a}

^{5}

=log

_{a}x=log

_{a}

^{3}Ö(9)

^{2}+log

_{a}

^{2}

=anti-log

_{a}x=anti-log

_{a}

^{3}Ö(9

^{2})+ log

_{a}

^{2}

=log

_{54}

x=54

c)log

_{b}(x-+3)=log

_{b}8-log

_{b}2

=anti-log

_{b}(x-+3)=anti-log

_{b}8-log

_{b}2

x+3=4

x=1

This is where my notes get messy and I'm really unsure which is question and which is answer so I'll put it down as I see it.

d)log

_{b}(x

^{2}+7)=2/3log

_{b}64

log

_{b}(x

^{2}+7)=2/3log

_{b}64

^{(2/3)}

anti-log

_{b}(x

^{2}+7)=anti-log

_{b}16

x

^{2}+7=16

x

^{2}=9

x=-3, 3

Also another note you can check your answer by filling in for x

e)log

_{b}x-log

_{b}(x-5)=log

_{a}

log

_{b}(x/(x-5))=log

_{a}

anti-log

_{b}(x/(x-5))=anti-log

_{a}

x=6(x-5)

x=6x-30

x=5

f)log

_{a}(3x+5)-log

_{a}(x+5)=log

_{a}8

anti-log

_{a}(3x+5)-anti-log

_{a}(x+5)=anti-log

_{a}8

2x+5=8x+40

-35=5x

-7=x

I'm not sure if these are from the questions he gave us I just can't remember, but from what I see I think you would have to change the whole number into a logarthym and then you use the anti-log to find x

a)log

_{2}(x

^{2})=4

log

_{2}(x

^{2})=log

_{2}16

anti-log

_{2}(x

^{2})=anti-log

_{2}16

x=-5, 5

b)log

_{3}6(x+2)+log

_{3}6=3

log

_{3}(6x+12)=3

log

_{3}(6x+12)=log

_{3}27

anti-log

_{3}(6x+12)=anti-log

_{3}27

x=5/2

c)log

_{6}(x+1)+log

_{6}=1

log

_{6}(x+1)+log

_{6}x=log

_{0}1

anti-log

_{6}(x+1)+anti-log

_{6}x=anti-log

_{0}1

\ no solution

Sorry for the long wait I just felt like doing the blog on a monday well it's a tuesday now but I started on monday If you see any mistakes feel free to tell me cuz I did it pretty late there's bound to be lot's of mistakes though, especially the last question. For the scribe umm... does anyone remember beat it by micheal jackson remember near the end where everyone crowded these two thugs and those two thugs had a weapon in one hand while the free hand was holding on to the other opponets free hand and the game was to hurt the other guy... well since there is marc and jessica left I believe they should do that to see who would be the scribe...winner doesn't have to do be scribe

## 1 Comments:

This is a good post. You chose to use colour to help illustrate you topic. Some of the colours you chose do not stand up (in the middle some of the browns and yellows look the same as the black font).

I think students in your class would understand the lesson as you presented it in your scribe. Is it hall of fame worthy?? YOu are close. Fix the colours in the middle and try to add something extra. COuld you find a link the explains your scribe. A tutorial on logs, practice quiz etc. This topic does not lend itself to colourful paint projects. Take it the extra mile and you will be in the Hall of Fame.

Mr Harbeck

Sargent Park School

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