Problem 1
Only answer True when the statement is ALWAYS True.
The function f(x) = e power of x / (x squared − 1) is continuous on [2, 5].
Problem 2
If log(x) = 4 then log(x^2) is ....
Problem 3
log base 2 (3x) + log base 2 (2x) = 3, what is x?
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53 comments:
in question one what do u mean by continous on the point 2,5??
well for problem three, the result when you expand would be:
log base 2 (3x) + log base 2 (2x) = log base2(2)^3
cancel logs to get:
(3x)x(2x) = 8
:6x^2=8
x^2=8/6
x= square root of 8/6
=1.154
The first problem is is through because u have to take (ln)first u have to right the problem in a certain form : e^x =x^2-1
: ln x^2-1 = x
then : ln x^2-1 = ln x
take out ln : x^2-1 = x
now if u substitute 2 in that equation u will get 5 so it countinous.(true)
PROBLEM 3
Dante i believe your result is incorrect.u basically have the right idea but i believe u got mixed up where u have log base 2 (2)^3.
This is the correct steps in order to work out the problem.
log base 2 (3x)+ log base 2 (2x)=3
log base 2 6x^2 =3
rearranged the equation
2 log base 2 (6x)= 3
log base 2 (6x)=3/2
put the equation into the exponential form and solve
2^3/2 = 6x
2.82 = 6x
x= 2.82/6
x=.47
PROBLEM 2
if logx = 4 then logx^2 = 8
because log x^ 2= 2 log x
since log x=4 then 2 log x = 8
With respect to problem 1 im a bit confused. If u substitute x into f (x) u wud get e^2 / 3. But wats next?????? copy cat!!!! how ln come into this?????
lookin at problem we can work out log(x) = 4 by solving for x which wud be x = 10^4 = 10000. If x is equal to 10000 then log(x^2) is log 10^8 = 8. I tink!!! Remember log 10^8 can be expressed as 8log 10. Wat u all tink???
For question 3 im abit confused… do I have to bring the term “3” to base 2???? Wat if I just apply the laws on the left hand side
log base 2 (3x) + log base 2 (2x) = 3,
(3x) (2x) = 3
6x^2 = 3
X^2 = 1/2
X = 1/4
Wats wrong wit dat?????
for problem 3 logbase2(3x)+logbase2(2x)=logbase2(2)^3??
I am stuck
to n2o
i a little confuse for your calculation on example 3
for problem 2, i think that the solution to this is that the log of x is to the base, therefore:
log[10]x=4, x^4=10
so log[10] x^2 : 2log[10]x
i agree with dante' because log x^2 = 2 log x
for problem 3 n2o is correct. u guys have to remember the laws and to be observant, because a common error is not identifing like bases (eg)problem 3.
in problem 2
log x = 4
log x^2= 2log x= 2(4) = 8
i'm a bit confused about question 1 what's really going on there?
is it that the question is asking whether it is true or false that e^x/(x^2-1) is continuos on [2,5]
for problem 2 zoe for some reason i feel like your missing something in that equation, double check and see what it is
can sum 1 explain number 1. wat does continous suppose to mean. tel me sumting. so i cud get rid of e by taking ln...jus as i wud take logs to get rid of powers. Right???
so if i cud get rid of e by taking ln how wud this apply for this question?
The number of milligrams of a drug in a persons system after t hours is given by the function
D = 20e^-0.4t.
When will the amount of the drug be 0.1 milligram (or almost completely gone from the system)? TRY DIS PLEASE??? IT CUD COME FOR XAM!!!!!!!!!!SO I CUD TAKE ln HERE?????????
i have a question:
differentiate this please
y = 4x^2 + log[3x^2]+2x^2 to base (3x).
N2O, I THINK THAT THE METHOD YOU WAS USING WAS INCORRECT. COULD SOMEONE ACTUALLY DO THIS?:
2 log base 2 (6x)= 3
LOG BASE 2 (6x) = 3/2
ALLYUH, THIS CORRECT?
cojebaby i dont know where u got the 8 from in your 2nd comment. can u explain further
like rat said wat do u mean by continuous on (2,5)
i dont understansd problem 1 completely would some1 shed som light into this for me please.
for problem 3 im tryin ah thig so if im wrong som1 correct me. our rule say log (a)(b) is the same as log(a)+ log(b).there4 rewrite the question like this so u would get
log base 2 (3x)(2x)=3
log (3x)(2x)/log2 =3
the logs would cancel of leaving (3x)(2x)/2 =3
6x^2/2=3
6x^2=3*2
6x^2=6
6x=square root of 6
x= d square rootof 6 / 6
is this correct any1
Dante's aswer for no. 3 is correct
he just skipped a part and confused you all.
log base 2 (3x) + log base 2 (2x) = 3
using:
log a + log b = log (a*b)
log base 2 (3x * 2x) = 3
log base 2 (6x^2) = 3
now using:
log base a (b) = c
b = a^c
a=2....b= 6x^2... c=3
2^3 = 6x^2
8 = 6x^2
8/6 = x^2
x = square root of 1.33
x = 1.155
n2o you mixed up the way you were supposed to eliminate.
here is what is really being done
8 = 6x^2
square root both sides to eliminate squared
8 square rooted = 6 square rooted * x^2 square rooted
2.828 = 2.449x
x = 2.828/2.449
x = 1.155
i dont understan problem 1 prob 2 log(x)= 4
2 log (x)= (4*2)
prob3
log<2> (3x)+ log <2> ( 2x) = 3
log <2> (3x*2x ) = 3
log <2> (6x^2)=3
is dat correct????
to pussinboots:
log(x)= 4 is the same as 10^4 = x
and log (x^2)= [10^4]^2
using the riules of indices we multiply the powers. 2*4=8
giving 10^8
i agree with dante' because log x^2 = 2 log x but for problem 3 logbase2(3x)+logbase2(2x)=logbase2(2)^3??
I stick dey
AYE!!!! nobody talkin bout number 2 again? i agree with n2o earlier on when he/she said that "logx = 4 ,logx^2 = 8... but there isnt any base stated in the question so won't it be base ten?
No:3
log base 2 (3x) + log base 2 (2x) = log base2(2)^3
drop logs to get:
(3x)x(2x) = 8
6x^2=8
x^2=8/6
x= square root of 8/6
=1.154
logx = 4 then logx^2 = 8
log x^ 2= 2 log x
since log x=4 then 2 log x = 8
problem 2 is 8 because if log(x)=4
then you take the 2 and put it to multiply so u will get 8
DANTE!!! Il have a shot at ur question y = 4x^2 + log[3x^2]+2x^2 to base (3x).
For this question the first ting i wud have done was to differentiate log[3x^2]to base 3x
so u = 3x^2
du/dx = 6x
dy/dx = 3x^2 / 6x log e to base 3x
dats the first part. Everybody understand that?
next just differentiate the remaining polynomials. the differential of 4x^2 wud be 8x and the differential of 2x^2 wud be 4x.
therefore my final answer wud be
dy /dx = 4x^2 + 3x^2 / 6x log e to base 3x + 4x.
Please correct me if im wrong.
PUSSINBOOTS!!!!! if x = 10000, then x^2 = 100000000 or 10^8. TRUE???? so lukin at the question log 10^8 = 8. remember 8 is the power and the base is understood as 10.
As if u didnt know log 10^8 is the same as log 10^8 to base 10. SAME TING!!! From the laws of logs
log a to base a is 1. therefore log 10 to base 10 is 1. as arule in logs u can bring the power as the coefficient. Understand????
can anyone answer this question:The current price of a litre of gasoline is $0.92 and is expected to increase at a rate of 12% every six months.The price of a litre of diesel is $0.80 and is expected to increase at a rate of 15% every 4 months. If these trends continue, after how many months will both fuels have the same price per litre?: The current price of a litre of gasoline is $0.92 and is expected to increase at a rate of 12% every six months.The price of a litre of diesel is $0.80 and is expected to increase at a rate of 15% every 4 months. If these trends continue, after how many months will both fuels have the same price per litre?
for problem 1,substituting x into f(x), i think that uu would get e^2/3, not quite sure, could somebody clear it up.....
for problem 2,
log(x)=4
then x=10^4
=10000
log(x^2) = 2(logx)
= 2(4)
= 8
problem 3
log base 2(3x)+ log base2(2x)=3
taking logs on both sides,
then the log would cancel off
3x+2x=3
5x=3
x=3/5
is this correct, can some1 please clear it up.....thnks
i think question 3 has been answered in a previous post on the blog. is it that peole are not learning from the blog
i am a bit confused with problem 2
can anyone varify the answer please cause no body is really thinking it out they are just diving into the problem
here is somthing to think about
why couldnt the answer be this:
since log(x) = 4
then log(x^2) = 4^2
= 16
??????????????????
no one is answering my question this is a logs question that i found.
R2D2:
I did some research with some UTT students and we came up with an idea. We decided to draw a graph. On the x axis we put the time and we put the prices on the y axis. We found the new gas and diesel price and plotted it with respect to 6 months. We kept finding the new prices until our graph intersected. This would give the time both gas and diesel would have the same price.
bootz:
since log(x)=4
the base is assumed to be base 10.
and since the base is ten. when we eliminate logs we will get:
10^4=x
This is because since 10 is the base in the logs equation and it is also the base in a the normal form then 4 would be the power.
Tazmania:
hope this will help u and myslef
logbase2(3x) + logbase2(2x) = 3
Then since both are adding we can multiply to eliminate one log.
logbase2 (3x * 2x) = 3
logbase2(6x^2) = 3
Then from here we can get rid of logs
6x^2 = 2^3
Then u can solve for x. Hope its correct
Bootz theres nothing hard about question 2 jus read my comment on question 2...u have to solve for x first..and find x^2 and use the log laws and solve. just read my comment. It makes no sense explaining this question again.
hey i, have a question, it real simple. differentiate:
y=2 Sin4x and
y= 45x Cos(3x+2)
For problem 3:
Log base 2(3x)+Log base 2(2x)=3
[log(3x)x(2x)]=3
Log base2 6x=3
2^3=6x
8=6x
x=8/6
x=4/3
x=1.33
someone tell if this is correct.
TO DANTE
Dats very simple
y=2 Sin4x
dy/dx = 8cos 4x
y= 45x Cos(3x+2)
dy/dx = -135x sin (3x+2)
Was i suppose to show any working. u said it was simple....didnt u????
in question 2 if logx=4 then logx^2 is 2 multiplied by for which is 8
for problem 3
log base2(3x)+log base2(2x)=3
=log base2(6x squared)=3
=2 cubed=6x squared
8=6x squared
x squared=6/8
x= the square root of 6/8
problem 3
log base2(3x)+log base2(2x)=3
=log base2(6x squared)=3
=2 cubed=6x squared
x squared=6/8
x= the square root of 6
2nd question
logx = 4
logx^2 = 8
log x^ 2= 2 log x
log x=4
2 log x = 8
Is this correct I think I did something wrong cam some1 plz help
Can some1 plz help me wit question 1 what is ment by continuous at
[2, 5].
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