Differentiate both sides??? N2o why wud u want to something like dat. It makes no sense. Since two out of the three terms are of base 10 it wud make sense to bring 6 to base 10. Ent??? so u wud end np with log (2x^2 + 6x) -6log 10 = log (2x) you can then use ur laws from there on...Differentiate???? dont even tink about it!!!!!
With respect to problem 2, again 2 out of three terms are of base 10. the ideal ting to do is bring 2 to base 10. so u wud get log (x + 6) = 2log 10 - log (4x. i wud say YES u are goin correct. BUT WHATS NEXT????? Any suggestions!!!!! anyone!!!!!!
1.)hmmmmmmm...analysing.... i think that u put everything in one suit.ie.put everything to the base 10.so log(2x^2 + 6x) - 6*10log 10(where *10 log 10 is 1 N.B.the symbol * means that the number is subscripted). then since all have the same base i think u take out all the logs. so 2^x +6 / 10^6 = 2x.
i tink hellion did a great job at explaining question 2 because i didnt agree with wat n20 wrote about differentiating the equation... u can onli differentiate if there are 2 unknown variables in the equation ... so if it was y= log 2x- log(2x^2 + 6x) -6... then dy/dx could be found..
For problem 1!!! wat if i brought all logs on one side..n know like log (2x^2 + 6x) - 6 = log (2x) log (2x^2 + 6x)-log (2x) = 6 thenuse the laws to solve from there. is dat correct?????
In problem one u can start by taking logs by the -6 so that the entire equation is possible to solve. After u have to bring all terms to the base ten so after u can use the rules of logs that is convert (-) to (/) on the left hand side of the equation. Then u can drop logs because all terms is now base ten. After u can easily solve for x.
well according to cokebaby and the gang bring all the logs on one side then we havelog(2x^2+6x)-log(2x)=6 log(2x^2+4x)=6 log(2x^2+4x)=log1000000 (2x^2+4x)=1000000 2(x^2+2x)=1000000 x^2+2x=500000 x(x+2)=500000 Therefore x=500000 OR 499998. This looking kinda crazy but it is what i arrive at following your advice . AM I WRONG!!!!!!!
well,lilo after thr 4th line i brought everthing to base 10. and if you log 1000000you will get 6. t dont know if it is right, and if it isn't i hope someone corrects me.
Where exactly in this question do you have to or say you have to differentiate? I think personally that no differentiation is invloved because they didn't ask for a gradient nor they didn't ask to differentiate.
Cokebaby makes a good point on how to solve this question. Since all the logs are to the base ten why not change the number six to the base ten then simplify from there. It looks more logical to do that from my point of view.
Working out question 1 you need to put like terms together in the equation, when you do that you should get log2x^+6x-log2x=6 then you take out the common term and then simplify .
well, from them comments, you have to simplify the equation first before solving. so the rules of logs applied in the question. and once everything to the same base and thing, you could use simplifications.
jason501 and everyone,please correct me if i wrong but,if you are logging one side you have to do it to the otherside also,because you can only take logs if the is no logs.this what i mean if you had 8x^2-9x=2x you carry across the -9x and then you can take logs,but if the question already has logs,you cannot just take logs for the six.
TO ALL THOSE WHO ARE DIFFERENTIATING, U ARE HEADING DOWN THE WRONG PATH BECAUSE ARE NOT ASKED TO FIND THE GRADIENT AT A POINT OR EVEN THE RATE OF CHANGE, ARE U?.. u start off by putting all the log entities on one side of the equation and the intergers on the other side. u will then get log (2x^2 + 6x)- log (2x)=6... then u sub 6=log10^6..which is 6log10
for problem 2 i believe ur on the right track but ur still far away from getting the answer...after this step, u take: log (x + 6)+log (4x)=log100 then u may know that log(xy)=logx + logy..so log(x + 6)(4x)=log100 log(4x^2 + 24x)=log100 4x^2 + 24x = 2 4x^2 + 24x - 2= 0 then u solve the equation using the quadratic formula...if u dont no this dont even bother trying differentiation
sly please explain further as to what u mean by " you can only take logs if there is no logs",u know, so i can help u.
TO ALL THOSE WHO SUGGESTED USING DIFFERENTIATION, EXPLAIN TO ME WHY U CHOSE DA ROAD.
I must commend spaceboy, u on d right track,just be a bit more worded in ur explanations please. he went right by first putting the like terms on one side,but someone please!!!!!anyone tell me which part LOG E come out from maybe i wasn't paying attention in class i real sorry for that but please i real lost.HELP.
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54 comments:
believe the working is:-
log(2x^2 + 6x)-6=log 2x
differentiate both sides an get it in terms of loge
so it would be
4x + 6/2x^2 +6x loge =2/2x loge
correct me if i'm wrong but i believe the working for equation one is:-
log(2x^2 + 6x)-6=log 2x
differentiate both sides an get it in terms of loge
so it would be
4x + 6/2x^2 +6x loge =2/2x loge
put everything on one side
4x + 6/2x^2 +6x loge - 2/2x loge
loge 4x+6/2x^2 +6x divide by 2/2x
loge 4x+6/2x^2 +6x multiple 2x/2
log e 2(4x +6)/ 2x(2x^2 + 6x)
log e 8x +12/4x^3 +12x^2
in example 2 i believe u are going wrong. This is what i believe to be correct.
log (x+6) = 2-log (4x)
log (x+6) + log (4x)=2
log (x+6)(4x)=2
4x^2 + 24x loge =2
differentiate
8x +24/4x^2 +24x log e= 2
cross multiple
8x+24/8x^2 +48x loge
can someone really explain this better for me
hey n2o where did the loge come from?
i agree with n2o up until he said loge. where did he get log e from?
Differentiate both sides??? N2o why wud u want to something like dat. It makes no sense. Since two out of the three terms are of base 10 it wud make sense to bring 6 to base 10. Ent??? so u wud end np with
log (2x^2 + 6x) -6log 10 = log (2x)
you can then use ur laws from there on...Differentiate???? dont even tink about it!!!!!
With respect to problem 2, again 2 out of three terms are of base 10. the ideal ting to do is bring 2 to base 10. so u wud get
log (x + 6) = 2log 10 - log (4x. i wud say YES u are goin correct. BUT WHATS NEXT????? Any suggestions!!!!! anyone!!!!!!
yes i think problem 2 was goin correct but i need a little time to think bout it.
i think you will have to bring the values in brackets to loge like what n2o said, but it hav a little more to the calculations.
1.)hmmmmmmm...analysing....
i think that u put everything in one suit.ie.put everything to the base 10.so log(2x^2 + 6x) - 6*10log 10(where *10 log 10 is 1 N.B.the symbol * means that the number is subscripted). then since all have the same base i think u take out all the logs.
so 2^x +6 / 10^6 = 2x.
in the first problem:
log (2x^2 + 6x) - 6 = log(2x)
rbirng all like terms to one side
log (2x^2 + 6x) - log (2x) = 6
laws of logs say:
log a - log b = log (a/b)
so:
log [(2x^2 + 6x)/(2x)] =6
pull out the common factor 2x
log [2x(x+3)/ (2x)]= 6
simplify
log (x+3) = 6
10^6 = x + 3
10^6 - 3 = x
i tink hellion did a great job at explaining question 2 because i didnt agree with wat n20 wrote about differentiating the equation... u can onli differentiate if there are 2 unknown variables in the equation ... so if it was y= log 2x- log(2x^2 + 6x) -6... then dy/dx could be found..
log (x+6) = 2-log (4x)
log (x+6) + log (4x)=2
log (x+6)(4x)=2
4x^2 + 24x log e = 2
8x +24/4x^2 +24x log e= 2
8x+24/8x^2 +48x log e
log(2x^2 + 6x)-6=log 2x
4x + 6/2x^2 +6x loge =2/2x log e
for number 1 use
log(2x^2 + 6)-6=log(2x)
in oreder to bring the 6 to a log form in the second part just write it like this
log(2x^2 + 6)-log base 10 to 10^6) = log(2x)
therefore when you bring all to a log form it wud be
log(2x^2 + 6)/10^6= log 2x
now u can cancel out the logs at both sides and solve
If im going wrong u can correct me. anyone i want to know my mistakes
hey hellion
i have to commend you on explaining the first question do you have any suggestions for the second question i think i need help.
For problem 1!!! wat if i brought all logs on one side..n know like
log (2x^2 + 6x) - 6 = log (2x)
log (2x^2 + 6x)-log (2x) = 6 thenuse the laws to solve from there. is dat correct?????
for problem 2: we assume that the log is to the base 10 but the i dont think that the base can be squared in that case
coke baby what you just said made alot of sense and up there with you on that but what is the next move waht laws are u talking about
In problem one u can start by taking logs by the -6 so that the entire equation is possible to solve. After u have to bring all terms to the base ten so after u can use the rules of logs that is convert (-) to (/) on the left hand side of the equation. Then u can drop logs because all terms is now base ten. After u can easily solve for x.
n2o you looking kinda wrong with that answer boy try dis
log(2x^2+6x)-log2x = 6
log(2x^2+6x)/2x = 6
10^6 = 2x^2+6x/2x
10^6(2x)= 2x^2+6x
hmmmmmm ah stick somebody finish it for me na
one
cokebaby meh bredrin finish the problem i put up na please
eh miss you gong correct in q2 because log2 is 100 like yuhself
i think problem two is going quite alrite so far.......
You should take it easycoke baby! it was a genuin mistake n2o made. Not everyone is as skilled as you.
And please be a little more specific cokebaby, exactly which of the laws you would use to solve.
well according to cokebaby and the gang bring all the logs on one side
then we havelog(2x^2+6x)-log(2x)=6
log(2x^2+4x)=6
log(2x^2+4x)=log1000000
(2x^2+4x)=1000000
2(x^2+2x)=1000000
x^2+2x=500000
x(x+2)=500000
Therefore x=500000 OR 499998.
This looking kinda crazy but it is what i arrive at following your advice .
AM I WRONG!!!!!!!
tinkerbell you kinda lost me there, where did all those big numbers come from
well,lilo after thr 4th line i brought everthing to base 10. and if you log 1000000you will get 6. t dont know if it is right, and if it isn't i hope someone corrects me.
so cokebaby u mean to say that 6log 10 is the same as 6... and from there u can easily say it is the same as log10^6??? ent dats wat u mean?
Where exactly in this question do you have to or say you have to differentiate? I think personally that no differentiation is invloved because they didn't ask for a gradient nor they didn't ask to differentiate.
Cokebaby makes a good point on how to solve this question. Since all the logs are to the base ten why not change the number six to the base ten then simplify from there. It looks more logical to do that from my point of view.
Working out question 1 you need to put like terms together in the equation, when you do that you should get log2x^+6x-log2x=6 then you take out the common term and then simplify .
an important thing to remember wen doning logs that contain brackets. you must log everything the within brackets
well, from them comments, you have to simplify the equation first before solving. so the rules of logs applied in the question. and once everything to the same base and thing, you could use simplifications.
for number one i think that you should just place the six on one side and work out the logs and then come back and to the six.Correct me if im wrong.
jason501 and everyone,please correct me if i wrong but,if you are logging one side you have to do it to the otherside also,because you can only take logs if the is no logs.this what i mean if you had 8x^2-9x=2x you carry across the -9x and then you can take logs,but if the question already has logs,you cannot just take logs for the six.
TO ALL THOSE WHO ARE DIFFERENTIATING, U ARE HEADING DOWN THE WRONG PATH BECAUSE ARE NOT ASKED TO FIND THE GRADIENT AT A POINT OR EVEN THE RATE OF CHANGE, ARE U?.. u start off by putting all the log entities on one side of the equation and the intergers on the other side. u will then get log (2x^2 + 6x)- log (2x)=6... then u sub 6=log10^6..which is 6log10
log (2x^2 + 6x)/2x = 6log10
log(x+3)=log(10^6)
so (x+3)=6
:. x=3
I hope everyone understands, if u r not sure substitute the val. of x in the equation and see what u get..
for problem 2 i believe ur on the right track but ur still far away from getting the answer...after this step, u take:
log (x + 6)+log (4x)=log100
then u may know that log(xy)=logx + logy..so
log(x + 6)(4x)=log100
log(4x^2 + 24x)=log100
4x^2 + 24x = 2
4x^2 + 24x - 2= 0
then u solve the equation using the quadratic formula...if u dont no this dont even bother trying differentiation
sly
please explain further as to what u mean by " you can only take logs if there is no logs",u know, so i can help u.
TO ALL THOSE WHO SUGGESTED USING DIFFERENTIATION, EXPLAIN TO ME WHY U CHOSE DA ROAD.
I must commend spaceboy, u on d right track,just be a bit more worded in ur explanations please.
he went right by first putting the like terms on one side,but someone please!!!!!anyone tell me which part LOG E come out from maybe i wasn't paying attention in class i real sorry for that but please i real lost.HELP.
ok u all r helpin me alot so i thank u but i dont konow bout the second problem. is it 2 equations or is the second one derived from the first one.
everybody askin bout dis differentation ting.. so y noe one is answering it ? coz i would really like to noe myslef
ok this is what i did.... im not too shure how correct it iz but here it iz:
log(2x^2+6x)-6=log(2x)
ok work out the brackets 1st
log(4x+6x)-6=log(2x)
log(10x)-6=log(2x)
1x-6=0.301x
group like terms
1x-0.301x=6
0.699x=6
cross multiply
x= 6/0.699
x=8.58
aye ratman how de jail u get dat.? first u differentiate wrong an second
u kar find de log of 2x...
i can work this question
3+ log base 2 5 =
3 log base 2 2 + log base 2 5=
log base 2 2^3 + log base 2 5=
log base 2 (8*5)
having said that.... can problem one be worked in that same manner.
log (2x^2 + 6x )- 6 log 10 =log 2x
log (2x^2 + 6x) - log 10^6 = log 2x
???????????
ratman please look at my previous comments, i suppose u r still a bit confused
problem 1:
log (2x^2 + 6x) - 6 = log(2x)
with all like terms on one side
log (2x^2 + 6x) - log (2x) = 6
by using laws of logs :
log a - log b = log (a/b)
log [(2x^2 + 6x)/log(2x)] =6
the common factor is 2x:
log [2x(x+3)/ (2x)]= 6
by simplification:
log (x+3) = 6
10^6 = x + 3
10^6 - 3 = x
concerning problem 2,
2 out of three terms are of base 10therefore the ideal ting to do is bring 2 to base 10.
this would give you:
log (x + 6) = 2log 10 - log (4x)
for all the confused ones out there the master has returned to help those in need. for the question
log (2x^2 + 6x) - 6 = log (2x)
all u have to do is rearrange the equation into a "better looking"
form as followes:
log(2x^2 + 6x)- log(2x) = 6
in doing this, u can then use the laws of logs to fuse the two log expressions into one as follows:
log[(2x^2 + 6x)/(2x)] = 6
in doing this we can now remove log to give:
(2x^2 + 6x)/2x = 10^6
which is equal to:
x + 3 = 1000000
and therefore
x = 1000000 - 3
x = 999997
hopes it clears up things
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