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Please if you know how to solve it, post the whole proccess of solving it, not just the result...

(the tasks are logarythms and exponential equation)

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Joined

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102 Posts

Please if you know how to solve it, post the whole proccess of solving it, not just the result...

(the tasks are logarythms and exponential equation)

And try in the future to solve your home work by your self . Or you might fail at the final exam .

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102 Posts

The problem is that I don't need a precise solution, I don't understand yours though, and the first equation was given to us for practise, there's nothing wrong with it...

Thanks anyway!

For the second one. Did you learn the "ln" ? (Sorry forgot it's scientific name) . If so, it's what I've used to solve it.

Sorry I tried my best, I told you it's been a while ^^

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102 Posts

and about the first equation, you must guess what x is, it can be only one solution, or an interval, so if you substitute x with 1, what you've proved is that the solution (x) isn't 1...

3 * (10^2x + 2*10^x) = (2*10^x + 1)^2

3*10^2X + 6*10^x = 4*10^2x + 4*10^x + 1

-1*10^2x + 2*10^x = 1

-2*ln10 + 2*ln 10 = 1 (Which is ofcourse wrong, since it's suppose to equal zero)

I recommend you to ask someone else ^_^ .

Anyways logs aren't too bad, they are just rules. Check out http://en.wikipedia.org/wiki/Logarithmic_identities to get a list of them.

Glup ^^" . Well maybe I should go check my old math lessons ^^" . And don't tell my old teacher

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102 Posts

Thanks a lot!

we finshed logs a week ago, and i bombed em lol

Anyway, I agree with him, there's nothing hard about logs.

yeh i must be retarded then, i found em hella hard....think i got 47% on the test...lolBoltzmann said:

Anyway, I agree with him, there's nothing hard about logs.

I've seen that too many people have problems with math because they're too lazy. It happens with my brother, and with my GF too. They complain that the stuff is too hard, but they aren't devoting their intelligence to it.Guyver2K5 said:yeh i must be retarded then, i found em hella hard....think i got 47% on the test...lol

Maybe that's your problem too

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102 Posts

Actually he only solved the first taskBoltzmann said:

Anyway, I agree with him, there's nothing hard about logs.

And about logarythms, they're not so hard, but our proffessor is very bad, he doen't know how to explain things very well that's why 25/26 students (!!!) failed the test...

The only person who was able to pass the test was a girl who was first on math championship (for the whole city), and she scored like 70%...

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15,879 Posts

I'll never forget our enthusiastic teacher do the 'log-wave' in front of the class

He might have some unorthodox teaching methods, but he's the best teacher of our school.

Just wait until you come to derivations and integrals. Once you are there, you'll wish back those simple equations. And you'll fall in love with the e^x function, trust me

Personally I feel that a lot of folks have a psychological barrier when it comes to this stuff and also a lot of other things in life. They convince themselves they won't understand it before they even try. Overcoming that is always the first step.Boltzmann said:I've seen that too many people have problems with math because they're too lazy. It happens with my brother, and with my GF too. They complain that the stuff is too hard, but they aren't devoting their intelligence to it.

Maybe that's your problem too

IMHO the best thing to do is stop looking at the large picture. You'll only convince yourself even more that the task is insurmountable. Break it down into little things that you know you can do, or at least attempt. Before you know it you've got the thing conquored and are left wondering what all the fuss was about.

Of course don't end up like yours truely, who is so busy that he effectively hasn't had a social life in over 5 years....

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Exercice 1:

Replace all the things by 0 and then abracadabra it appears that this is the solution

-10^2X + 2*10^X= 1 <-> -1+2=1 <-> 0=0

Exercice 2: the answer is x = ln(12)/ln(12/5)

5^X=12^X/12 <=> 5^X=12^(X-1) <=> x ln 5 = (x-1) ln 12

(x-1)/X = ln(5)/ln(12) <=> x = ln (12) / [ ln (12) - (ln5) ]

then You use ln a - ln b = ln (a/b) and you find my result

exercice 3: Quickly I found that it's X= 2^(1/4) * e ^(1/4) just used the fact that log (e)=1 and the formula log (a) + log (b) = log (ab) so the equation became log (2e*x^3) + 2 log (x) = 0 thus 2ex^4 must be equal to 1.

Thank you my brain was a bit rusty :hdbash: (<- the only way to start this f....ing machine,give me another one mine is defect)

But now my brain should work fine !

You just need to remeber a few things about logs.

Ln (x.y) = ln(x) + ln from this you deduct that ln(x²)=2 ln (x) etc... so

ln (x^n) = n*ln(x)

and ln(x/y)= ln(x)-ln so ln(1/x) = ln (1)-ln (x) = -ln(x)

ln(0) has nos solution and ln(e)=1

to go further: ln is the primitive of 1/X, e^x is a function defined to grow the fastest possible so it's derivative is e^x

You can manually calculate a good approximation of e by using the limited developpemnt (just traduced from french) e is infinite (in length)

e^1= SUM (1^k/k!,k=0,n)=1+1+1/2+1/6+1/24+1/120 remember 0! = 1 & 1!=1

& n!= *(n-1)*(n-2)...*1*1

Log & exponentiel are widely used cause they considerably simplify calculations, and describe sometimes perfectly natural phenomenom.

If 10^X = 1

-> X = log 1

-> X = 0

Check:

3(10^2X+2*10^X) = (2 *10^X + 1)^2

-> f(x) = 3(10^2X+2*10^X) - (2 *10^X + 1)^2

Where X = 0:

f(x) = 3(1+2) - (2 * 1 + 1)^2

f(x) = 9 - 3 ^ 2

f(x) = 0

Logs are a pain :/

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