for d 2nd question u are goin correct because lg of 100 can b written as lg 10^2 and since log base 10 by 10 =1 and removal of d powers to d front of the lg gives 2. so changing all d terms to lgs makes it easer
are we solving these questions as simultaneous equations.but when u're calculating the 2nd eqaution in problem2 won't u have 2 used the 2^3=8 approach to get x after all the knowns have been grouped and calculated.
I agree wid zipper ..but as miss always say wen u dont see de base remember is 10...so we put base 10 for the logs...we shud den place all the logs on one side and the number on the other..since it is the same base
For question 2..u are goin correct and i agree wid zipper 100%..lol...wat more can i say...log 100 is equal to 10^2 and it also give log base 10 100 which is log base 10 10^2..dat just makes it easier..as miss say .."if u see it can work out..work it out.." so it can
no 2 is absolutely correct because log 100 is actually 2. here the person has a strategy to have everything in log terms, so that it will be easier to work out!
QUESTION 1 One of the very first things we learnt in logs is that in the equation we place all logs on one side and values on the other. log(2x^2+6x)-log(2x)=6
as seen above all the logs are the same meaning they all have base 10.
log((2x^2+6x)/2x)=6 So i fully agree with zipper.
QUESTION 2 The log of 100 is equivelent to 10^2 which also gives log base 10 100 which is log base 10 10^2
wel i don't quite understand wat mysticwings said but if i had 2do ques 1 then i'd say dis..
log (2x^2 + 6x) - 6 = log (2x) wel firstly i'l multiply log wit everyting in d brackets so it read log 2x^2+log 6x-6=log 2x then we group al log terms log2x^2-6=log2x-log6x k as we c on d rite side its a minus so we divide,however the left side remains the same log2x^2-6=log2x/6x log2x^2-6=log3x then we brin al d log terms 2gether again log2x^2-log3x=-6 as the rules aplies wen its a minus u divide log(2x^2/3x)=-6 log(2x/3)=-6 since logs is base 10 it becomes 10^-6=2x/3 0.000001=2x/3 and wel u cross multiply u get d ans.
I will try to help you angel. From here, I think its this:
log (12x^3)/(2x)=6
When no base is seen, it is defaulted to base ten (10). So, we can now do change it back to the indices form. I think you should have written this in the beginning as miss always does:
2^3 = 8 log 2 base 8 = 3 or log 2^8 = 3. Its d same thing. Jus trying to explain.
Then you can now see the relationship to miss question.
for problem 1 log(2x^2+6x)-6=log2x u can place all the logs on one side of the = & the numbers on the other side => log((2x^2+6x)-log2x=6 now, when subtracting logs with the same base, we divide: follow the rule from then on, it's very simple
for question 2 log(x+6)= 2-log(4x) log(x+6)= log100-log(4x) you are absolutely going correct because the log of 2 is 100. ppl, if u don't agree, use ur calculators and punch in the figures!!! WAW!!!!!!!
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ok for problem 1:
ReplyDeletewe can bring all d logs on 1 side and leave the no. on the other..
so log(2x^2+6x)-log(2x)=6
since its d same base=
log((2x^2+6x)/2x)=6
for d 2nd question u are goin correct because lg of 100 can b written as lg 10^2 and since log base 10 by 10 =1 and removal of d powers to d front of the lg gives 2. so changing all d terms to lgs makes it easer
ReplyDeleteare we solving these questions as simultaneous equations.but when u're calculating the 2nd eqaution in problem2 won't u have 2 used the 2^3=8 approach to get x after all the knowns have been grouped and calculated.
ReplyDeleteI agree wid zipper ..but as miss always say wen u dont see de base remember is 10...so we put base 10 for the logs...we shud den place all the logs on one side and the number on the other..since it is the same base
ReplyDeleteFor question 2..u are goin correct and i agree wid zipper 100%..lol...wat more can i say...log 100 is equal to 10^2 and it also give log base 10 100 which is log base 10 10^2..dat just makes it easier..as miss say .."if u see it can work out..work it out.." so it can
ReplyDeleteCan someone please explaion question 1 in full details for me please
ReplyDeletefor Q1 i cant go further than zipper went.
ReplyDeleteall i kno is that log with no given base is understood as log base 10, but then isnt log base 10 = ln?
well boy-b ill explain no 1 to you:
ReplyDeletethe first thing to do is to group all the common terms together:
log (2x^2 + 6x) - log 2x = 6
from our log rules we know that when subtracting common logs, we can divide one log expression by the other:
log (2x^2+6x)/2x = 6
from here we simply, it just normal algebra:
log (x+3) = 6
now we can use antilog to find for x:
(x+3)= antilog 6
(x+3) = 1000000
x = 1000000 - 3
x = 999997
no 2 is absolutely correct because log 100 is actually 2. here the person has a strategy to have everything in log terms, so that it will be easier to work out!
ReplyDeleteQUESTION 1
ReplyDeleteOne of the very first things we learnt in logs is that in the equation we place all logs on one side and values on the other.
log(2x^2+6x)-log(2x)=6
as seen above all the logs are the same meaning they all have base 10.
log((2x^2+6x)/2x)=6
So i fully agree with zipper.
QUESTION 2
The log of 100 is equivelent to 10^2 which also gives log base 10 100 which is log base 10 10^2
This comment has been removed by the author.
ReplyDeletesince all the logs are to base 10 i would approach this question by placing all the logs on one side and placing all the logs on one side
ReplyDeletewel i don't quite understand wat mysticwings said but if i had 2do ques 1 then i'd say dis..
ReplyDeletelog (2x^2 + 6x) - 6 = log (2x)
wel firstly i'l multiply log wit everyting in d brackets so it read log 2x^2+log 6x-6=log 2x
then we group al log terms
log2x^2-6=log2x-log6x
k as we c on d rite side its a minus so we divide,however the left side remains the same
log2x^2-6=log2x/6x
log2x^2-6=log3x
then we brin al d log terms 2gether again
log2x^2-log3x=-6
as the rules aplies wen its a minus u divide
log(2x^2/3x)=-6
log(2x/3)=-6
since logs is base 10 it becomes
10^-6=2x/3
0.000001=2x/3
and wel u cross multiply u get d ans.
wel i asume the base of these questions are all 10. I HONESTLY WOULD NOT HAVE THOUGHT THAT UNTIL I READ SOME COMMENTS THANKS!!!
ReplyDeletetherefore i will do what zipper did and bring all the logs on one side and keep all the values on the other.
mysticwing i tink u made a mistak 6x/2x=3x.this is 4 ques.2
ReplyDeleteTherefor for question 1, log (2x^2 + 6x) - 6 = log (2x)
ReplyDeletei would get log (2x^2 + 6x) - log (2x) = 6
i will then expand the brackets and get
log (2x^2) + log (6x)- log (2x)= 6
i will then use the rule which states that when adding logs of the same base you must multiply the powers therefor i will get
log 12x^3 - log(2x)= 6
now ill use the rule which states that when subtracting logs of the same base you must divide the powers therefor i will get
log (12x^3)/(2x)=6
therefore you will get log(6x^2)=6
I LOST AGAIN PLZ HELP. I WILL GET IT ON MONDAY CAUSE I DONT HAVE ACCESS TO NET AT HOME.
i don't understand hw its 100 can sum1 explain caz i would hav used another way
ReplyDeleteFor problem 2 question two i really did not see that 10^2 would have been expressed as 100. I would have worked it out wrong.You sure that is correct.
ReplyDeleteI will try to help you angel. From here, I think its this:
ReplyDeletelog (12x^3)/(2x)=6
When no base is seen, it is defaulted to base ten (10). So, we can now do change it back to the indices form. I think you should have written this in the beginning as miss always does:
2^3 = 8
log 2 base 8 = 3 or log 2^8 = 3. Its d same thing. Jus trying to explain.
Then you can now see the relationship to miss question.
log(base10) (12x^3)/(2x)=6
Hence,
10^6 = (12x^3)/(2x)
1’000’000 = (12x^3)/(2x)
Then you can simplify and find for ‘x’.
That is a very good explaination Kurosaki Ichigo
ReplyDeletefor problem 1
ReplyDeletelog(2x^2+6x)-6=log2x
u can place all the logs on one side of the = & the numbers on the other side
=> log((2x^2+6x)-log2x=6
now, when subtracting logs with the same base, we divide: follow the rule
from then on, it's very simple
for question 2
ReplyDeletelog(x+6)= 2-log(4x)
log(x+6)= log100-log(4x)
you are absolutely going correct because the log of 2 is 100.
ppl, if u don't agree, use ur calculators and punch in the figures!!!
WAW!!!!!!!
Problem 1
ReplyDeletelog (2x^2 + 6x) - 6 = log (2x)
the first thing to do is to identify what base the logs have. Since none is wrote in the question base 18 is understood.
then the second thing to do is to put all the logs on one side of the equation
log (2x^2 + 6x) - log (2x) = 6
Since the logs are being minus we divide the logs.
log [10] (2x^2 + 6)/ 2x = 6
This can now be switched to exponent
10^6 = 2x^2 + 6 / 2x
Its just a matter now of solving for x
Thanks for the help, i finally got out how to solve question 1
ReplyDelete