Look at the following and made some comments to assist. Remember the log base is in [] brackets
Example 1
log [8] (x + 2) = 2 - log [8] 2
log [8] 1 + log [8] 2 = 2 / (x + 2)
Example 2
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
log (2x + 4) = 2
log (2x + 8) = 2
log 8
2x + 4 = 1.8
2x = 1.8 - 4
Example 3
log [8] (x + 2) = 2 - log [8] 2
log [8] x + log [8] 2 = 2 - log [8] 2
log [8] x + 2 log [8] 2 = 2
Example 4
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) = 1.67
log [8] 1.67 = x + 2
Example 5
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
log [8] 2(x + 2) = 2
log [8] (2x + 4) = 2
log [8] 2x = 2 - 4
Example 6
log [8] (x + 2) = 2 - log [8] 2
log (x + 2) / log 8 = 2 - log 2 / log 8
log (x + 2) / 0.903 = 2 - 0.33
log (x + 2) 0.903 * 1.667
log x + 2 = 1.5
log x = 1.5 - 2
log x = 0.5
x = 0.32
Subscribe to:
Post Comments (Atom)
73 comments:
log is a functon used to decrease the size of numbers?
i think that eg1 has an error 1+ is does not have to be there in log [8] 1 + log [8] 2 = 2 / (x + 2)
.Also in eg2 the corrected line in log (2x + 4) = 2 has to be log8(x + 4) = 2.
captin jack i realy do not understand your point please explain.
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
log [8] 8 =1 therefore 2log [8] 8=2
log [8] ((x + 2)x(2))=
log [8](8)Squared
((x + 2) x (2))=8 squared
2x + 4 = 64
2x = 60
x = 30
this is sort of a simplified form to solving the question.
i AGREE WITH SLEEPY'S EVALUATION, I DID IT SIMILAR INTHE EXAM
all the examples are wrong. why are these examples given? miss, by any chance is the examples given, some of the solutions that students gave.
Which one of theseexamples are correct?
I think they are are all wrong according to poison One of them ooks like what i did in exam .
example1; the second part wrong
example2; the forth part wrong
example3; I don't know wat going on dey
example 4; the third part is wrong
example 5; how do u get x?
example 6; how did the 1.667 become positive?, where the (/)sign by 0.903?, also how was x obtained?
All of the six examples give incorrect answers for x … the question states that log [8](x+2) = 2 – log [8]2 … it is fairly simple to calculate x… all that needs to be done first is change 2 in terms of logs as sleepy did… therefore u ask yourself log [8] of what would give the answer of 2…. log [8] 64= 2… now the equation becomes…log [8](x+2) = log [8] 64 – log [8]2 …then it is easier to solve .. log [8](x+2) = log [8] (64/2) … the logs would cancel an u would b left with x+2 = 32 … an x=30…
will i can see that all the examples are wrong. the main problems are not transposing properly, u need to be very carefully in terms of what u transfer and try to do it accuratelly. not like example 1!
i will now explain another question in a simplified manner.
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
then log[8] (2(x+2))= 2log[8]8
(because log[a] a = 1)
cross off the logs and simplify
2x+4= 8 squared
2x= 64-4
x=60/2
x=30
i agree with j joker. the examples are wrong. if you revise the rules of logs it must seem easier.
alright folks here it is
first group the log terms
log[8](x+2) + log[8]2 = 2
following the log rule of
log(MN)= log M+log N
we get
log[8](2{x+2})= 2
log[8]2x+4 = 2
so
8^2= 2x+4
so
64= 2x+4
so
x= 64-4/2
so
x=30
who say answer correct!!!!!!
after looking at the examples it was seen that the students employed the wrong method in hanlding the question. most of the comments just solved the problem and didnt explain how to work it. To the students having a problem with logs it is very important that u know the laws and know when to apply them. please know when to apply ur laws. in solving any log problem of this nature please note that the first ting u do is try to bring all the terms to some common base ok. from the example two out of the three terms are of same base 8. how wud u bring the term "2" to the base 8?? Remember from the laws.... log[a]a = 1..bring "2" to base 8 we wud get 2log[8]8. Make sense?? remember log[8]8 = 1. now u have all terms to the same base. log [8] (x + 2) = 2log[8]8 - log [8] 2
the next step again involves applying the laws.
log[a]b - log[a]c is the same as
log[a](b/c) applying this law to the equation:
log [8] (x + 2) = 2log[8]8 - log [8] 2....u would get
log [8](x + 2)= log[8](8[2]/2). Understand??. To solve just drop logs so u wud get,(x + 2) = 32. so x = 30. jus use the laws!!! i didnt know anyting about logs and i just learnt the laws and jus practice, practice, practice...
thanks sleepy I followed your technique for this same question
From reading sleepy's comments, I realised that there are a few rules in working logs that I am not too familia with so I am certain that I would have made one of these mistakes listed. I see that the issues in solving this probles lies within working with brackets and identifying like terms and linking them correctly.
what you need to remember is a simple stratagey for instance 2(3)=8
this means log(2)8 =3 see where these terms are changed bu it means the same thing but ina dirrerent form.
captain jack sparrow ur comment confuses me. i tink it does not necessarily decrease the size of a number.I ting logs simply is an alternative way of writing an index. for example 125 can be expressed as 5^3
HEY PEOPLE U ALL ARE NOT ANSWERING THE QUESTIONS RIGHT. MISS SAID TO COMMENT ON THEM NOT ONLY SAY IF IT IS WORNG
I THINK THE EXERCISE IS TO C HOW MUCH WE ALL UNDERSTAND THE LOGS
capt jack sparrow.....never heard of that function of logs. i know one function is to make curve line graphs into straight ones so that deductions can be made.
to captain jack sparrow's first comment:
a log is a function similar to a plus or divide. It does not make numbers larger or smaller, it merely expresses them differently. i hope that helps.
i'm a little confused, can somone explain example one
in example 1.. it is wrong and should be done....
log [8] (x + 2) = 2 - log [8] 2
log[8] (x+2) + log [8] 2 = 2
log[8] 2(x+2) = 2
8^2 = 2(x+2)
64 = 2x + 4
60 = 2x
x = 30
it is important to know what you can and cannot do with log questions. always remember the significance of the brackets in logs.. it is there for a reason.
hellion.. look at sleep's solutuion of example 1.
log [8] (x + 2) cannot be written as log [8]x +log[8]2.
this is a big error and should not be confused with expanding brackets normally as is done in
2(x+2) = 2x +2
wong fei i dont understand yur las comment what do u mean
hellion!!! u ought to be confused for example 1 cuz its wrong..u aint read my first comment oh wat??? read it and tell me wat exactly u dont understand??
i dont understand #6 completely. i understnd to the point up 2 the brackets disapper by the log on the right hand side.
ummm... pussinboots!!! u know question 6 is wrong rite? the method used to solve this problem was wrong..at least i tink so. any ever heard of dat method???
I notice in the third line of the statement in question 2, that the base 8 is removed or gone. when did that happen or how?
I think there, almost, maybe, miiiight be an error in example2 line 4, should't it be 'log(2x + 4)=2.and Where did the log go in line 6, did it get canceled with the log in the denominator?
Wait, hold up, these are examples with errors in them. I thought these examples where legit. woops! or are they? I got some serious work to do. Yo hellion could you answer the question i asked in the first section of this blog.
Ahhhgh,this question is directed to wong fei hong. in your answer to hellions statement you explained the proper answer to example one,right? so could you explain where the '8^2'came from in the 4th line. Just for clarity. Please.
crimson crock!!!wat wong fei hong did was that he used the rule... u ever hear about...
log[a]b=c can be expressed as
a[c]=b, so this law applies to dat part of the question.
log[8] 2(x+2) = 2
8^2 = 2(x+2)
Understand???
in example 4 would the answer be
log [8] (x+2)= 2-log [8] 2
draw a line
log (x+2)= 2 log 2(-1)
log [8] log [8]
drop logs
(x+2) = 0.5
4 8
cross multiple
8(x+2)= 2
8x + 16= 2
8x = 2-16
8x = -14
x =-14/8
x= -1.75
i'm not sure about that answer, if its wrong can some-one help explain it better
As i look through i see that all 6examples have some sort of errors in them, firstly we need to examine them and use a basic example of 2^3=8 =log2 8 =3 and from the the questions we compare to find for x.
I believe example 6 is correct when i work it out and the other examples are wrong.
i have never been to good at locating errors in an equation but by viewing every one's comments i think by now i have fair idea of what to look for
to ichigo:
no. 6 is wrong. Log (x+2) cannot be separted like that. log (x+2) is the logarithm of (x +2). Since x+2 will equal one value then it should be:
log (X+2)=1.5
10^1.5 = x+2
31.62 = x + 2
to hellion:
how do you which method to use. do the () have anything in determining the way to proceed in this question
to tweety:
the () show that the contents within the brackets are to be logged. If there were no brackets then the +2 would have
Wel in the quiz i didnt do this question because i didnt know how to start it,I was confused!
i the quiz i made an error in question 3 which was:
solve for the given equation
log[8](x+2) = 2-log[8]2
this is what i did:
log[8](x+2) = 2-log[8]2
log(x+2)/log8 = 2-log2-log8
log(x+2) + log8 + log 8 = 2
log(x+2) = 2
2x = 2-4
x = -2/2
x = -1
could someone please help/show me the right way in doing this question
Now I can say that for the examples what is right and what is wrong because i did sum revision.
For this question the common mistake is that time was not taken to understand the question most people just dove in and got stuck
Example1-Nice cross multiplication
Example2-second part wrong
Example3-Whats wrong with you,where is the bracket term.
Example4-WRONG!!!!!!
Example5-Looks good but u kinda got me confused there.
Example6-Wrong.
the first ting one must note is 2 is not to the base 8 as the other expressions so the first ting one should do is to get an expression to the base 8 and according to sleepys caluations he is correct
Captin what do you mean by that FIRST comment.EXPLAIN!!!!!!!PLEASE!!!!!
log[8](x+2) = 2-log[8]2
carry logs on one side
log[8](x+2) + log[8]2 = 2
Usinglog[a]b + log[a]c = log[a](bc)
log[8]2(x+2) = 2
log[8]2x+4 = 2
usingx=log[a]y-->y=a^x
8^2= 2x+4
64=2x+4
64-4=2x
60=2x therefore x=60/
=30
PRACTICE YOUR LAWS
That last comment was long and as everyone knows that all the examples are incorrect. Therefore its quite simple,put log terms together and simply equate.
TO SLEEPY:
You are quite a brillant person and i must say we have to meet.LOL.As for the logs at least i get it now or i would have had to come to you but maybe i still will,if thats ok.KEEP UP THE GOOD WORK!!!!!!!!!!
after trying back these examples there is only one way to do this
i'm still afraid to remove the brackets when it comes to logs ..... what rules apply when removing such brackets and when it should be removed
For question 5 in the quiz which is not posted
5^2x-1 = 6^x+3
I just understood it a while ago.
here is what i came up with
log 5^2x-1 = log 6^x+3
2x-1 log 5 = x+3 log 6
2x-1/x+3 = log6/log5
2x-1/x+3 = 1.11
2x-1 = 1.11(x+3)
2x-1 = 1.11x + 3.33
2x-1.11x = 3.33 + 1
0.89x = 4.33
x = 4.33/0.89
x = 4.86
Even though this question was already answered thanks to MOOSE i now know how to answer it.
So with my new knowledge is anyone willing to give me a question to solve????
3^x = 10^2x+7
bootz work this one out for meh yuh bake yuh should ah never!!!!!
good luck hope you get it out put it up so i could see it na
from the exam i realised that a friend still didn't know how to convert an equation from logs form to exponential form, for those who are still confused:
log [x]a=y where x is the base, y is the power and a is the solution to or the answer of x to the power of y.
this can be written as [x]y=a
inexample5you were going fine until
u wrote log[8]2x=4-2
You need to go back to your rule and see how the problem is expressed.Using y=a^x---> x=log[a]y
you should have written 2x+4 = 8^2
So 64 = 2x + 4
64-4 = 2x
therefore x = 60/2
=30
For this question,you could group all the like terms together and then equate to start solving for x;so then the next step should be:log[8](x+2)+ log[8]2=2. Another error someone can make is:
(x+2)log 8= 2-[log 2/log 8].
log[8](x+2) = 2-log [8]2
ok here goes
log[8](x+2) + log [8]2 = 2
log[8][(x+2)(2)] = 2
log[8](2x+4) = 2
8^2 = 2x+4
64 = 2x+4
64 - 4 =2x
60 = 2x
x = 60/2
x = 30
yay now i understand!!!!!!!!!!
looking at example 2 where did
2x + 8 come from you need to take your time when writing down your stuff
BE CONSISTENT....Take your time
is this correct moose??????
3^x = 10 ^2x+7
log 3^x = log 10^2x+7
xlog 3 = 2x+7 log10
x/2x+7 = log10/log3
x/2x+7 = 2.1
x = 2.1(2x+7)
x = 4.2x + 14.7
4.2x - x = 14.7
3.2x = 14.7
x = 14.7/3.2
x = 4.6
The error in question one lies in two places. 1. log[8](x+2) is one term and cannot be expanded.
i have a question...how can i express the equation y = ax[n] in the form y = mx + c????
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
log [8] 8 =1
2log [8] 8=2
log [8] ((x + 2)x(2))=
log [8](8)^2
((x + 2) x (2))=8^2
2x + 4 = 64
2x = 60
x = 30
eh bootz yuh very correct yuh now catchin d drift of logs
The Solution to question 1 is as follows:
log[8] (x+2) = 2-log[8]2
Now using a nifty trick my lecturer showed me we get the following:
log x+2/log8 = 2- log2/log 8
Simplifying a bit we get:
logx+2/ 0.903 = 2-0.33
Further simplfication:
log x+2/ 0.903 = 1.67
Now we multiply both sides by 0.903
log x+2 = (0.903)(1.67)
log x+2 = 1.5
x+2= log^-1 (1.5)
x+2= 32
x= 32-2
x= 30
To my understanding in example 2 after the -log[8] 2 was transposed,
the 2 was miltiplied by the bracket. If that was the case then that is a big error. Then the log base 8 just evaporate. That question real wrong.
prob 1
log bse 8 (x+2) = 2 log bse 8(8) - log base 8 2
log<8> (x+2) = log<8> 8^2 / log<8>2
the correction for prob 2 supposed to be log <8> ( 2x + 4) = 2
the correction for prob 3
log <8>x + log<8> 2 = log <8> 8^2 - log <8> 2
Post a Comment