| Autor |
Beitrag |
    
Christine
| | Veröffentlicht am Montag, den 15. November, 1999 - 15:09: | 
|
Hallo
Da ich krank bin und heute nicht in die Schule gehen konnte, weiß ich nicht wie man folgende
Aufgaben löst. Bitte helft mir!
1) 2 hoch x = 3 hoch (x-1)
2) 7 hoch (x+1) = 2 hoch 7x
3) 5 hoch 2y = 4 hoch (1-y)
4) 4 hoch (2z+1) = 10 hoch 3z
5) 3 mal 1,4 hoch 3t = 2 hoch (t-1)
6) 4 mal 5 hoch (x-1) = 10 hoch (x+1)
7) 7 mal 6 hoch 2x = 11 hoch (x+3)
8) (3/4) hoch (3x-2) = (2/5) hoch (2x-3)
DANKE!
P.S.: Schön wäre, wenn der genaue Rechenweg dabei ist. |
Sven
| | Veröffentlicht am Montag, den 15. November, 1999 - 16:54: |
|
Hallo Christine,
ich habe gerade leider nicht sehr viel Zeit, aber einen kleinen Tip
log(a hoch b) = b * log a
lg ist log zur Basis 10
Lösung zur 1. Aufgabe auf beiden seiten lg anwenden
=> mit Tip
x * lg2 = (x-1)*lg3 nach x auf lösen
x(lg2-lg3) = -1*lg3
=> x = ... |
Tina
| | Veröffentlicht am Dienstag, den 30. November, 1999 - 18:11: |
|
Hilfe, schreibe morgen ne mathearbeit über Exponentialfunktionen und kreige das nich gepeilt |
Bodo
| | Veröffentlicht am Dienstag, den 30. November, 1999 - 22:12: |
|
Schau Dir am besten Beispielaufgaben zu dem Thema hier im Board an (gibt ne Menge).
Oder: Was ist Deine Frage?
Bodo |
Sushi
| | Veröffentlicht am Donnerstag, den 13. Januar, 2000 - 12:18: |
|
Hallöchen,
ich schreibe morgen auch eine Mathearbeit über Exp.-Funktionen.Den Rechenweg oder Tip von Sven verstehe ich aber absolut nicht...
Ich habe auch etwas andere Aufgaben:
(2a - 5b)hoch 6 durch (4a - 10b)hoch 6
144 durch (240 hoch 3 minus 7 hoch 4)
Wäre nett wenn mir damit jemand noch heute (verständlich) helfen könnte!!! |
Karo
| | Veröffentlicht am Sonntag, den 23. Januar, 2000 - 15:50: |
|
Hallo ich bin karo!
Und habe da so eine frage weil ich bald ne scheiß (sorry) mathearbeit schreib!!
Also bitte bitte bitte HILFE!!!!!!
3hoch3x+2 + 7hoch2x-1 =3hoch3x-2 + 7hoch2x+1
also bitte ich brauch den rechenvorgang!!!!!
danke!!!
PS bitte so schnell es geht |
Sven
| | Veröffentlicht am Montag, den 24. Januar, 2000 - 20:01: |
|
Also erstmal etwas lesbarer:
33x+2+72x-1=33x-2+72x+1 <=> 33x+2-33x-2=72x+1-72x-1 <=> 33x-2(34-1)=72x-1(72-1) <=> 80*33x-2=48*72x-1
Jetzt logarithmiere beide Seiten: Z.B. gilt: ln(33x-2)=(3x-2)*ln3
Dann kannst Du zusammenfassen und nach x auflösen.
Noch Fragen? |
Karo
| | Veröffentlicht am Mittwoch, den 26. Januar, 2000 - 19:15: |
|
Sven ich danke dir!!!! Bist ein Schatz!!!!
Ciao Karo!!!BUSSI |
Jan
| | Veröffentlicht am Dienstag, den 01. Februar, 2000 - 01:12: |
|
Hey,
I've got a question according to the following problem:
(-4^-2)-1
the answer is -16 right? Could you prove it mathematically? Why is it -16 and not +16
and if it's possible, I need more than "the exponent doesn't affect the negative sign". Is there a mathematical way to prove it besides that rule? Please help me I would really appreciate it, thanks a lot!!!
Jan!!!
PS: If you don't mind, please explain it in English! |
Jan
| | Veröffentlicht am Dienstag, den 01. Februar, 2000 - 03:59: |
|
Hey,
I gotta correct the problem.
It is
(-4^-2)^-1
Could anybody explain it to me (please regard my other mail!)
Thanks,
Jan!!! |
Fern
| | Veröffentlicht am Dienstag, den 01. Februar, 2000 - 09:27: |
|
Hello Jan,
Negative exponents can be converted into positive ones by putting the term into the denominator:
a-p=1/ap
In your expression it is not clear whether you mean:
(-4)-2
or
-(4)-2
In the first case we have:
(-4)-2 = 1/(-4)2 = 1/16
This expression in turn is to the power of (-1):
(1/16)-1 = 1/(1/16) = 16
================================
If you mean: -(4)-2=
-1/4² = -1/16
and again:
(-1/16)-1=-1/(1/16)= -16
============================== |
Jan
| | Veröffentlicht am Mittwoch, den 02. Februar, 2000 - 03:07: |
|
Hey,
first of all, thanks for your effort
I'm sorry to disturb you again,
I mean:
(-4^-2)^-1
(say):
put in parantheses: negative four to the negative second, the result of the parantheses to the negative one
Let me tell u what I think:
The exponent negative two in the parantheses doesn't affect the negative sign of the four, since the even exponent doesn't affect the negative sign when there are no parantheses
{ (-a)^2 would not be the same as -a^2 }
so the result of the parantheses is -(1/16) and this to the -1 gives you -16
my question (which is probably kind of weird) is (if my explanation is right), if there is a mathematical proof that -a^2 is not the same as (-a)^2 (and that the answer to my problem is -16 and not 16 because of this reason)
I only know that rule, that it doesn't affect the negative sign without parantheses.
I hope I could express myself better this time,
thank u a lot again!!!
Jan!!! |
Jan
| | Veröffentlicht am Mittwoch, den 02. Februar, 2000 - 03:09: |
|
Hey,
first of all, thanks for your effort
I'm sorry to disturb you again,
I mean:
(-4^-2)^-1
(say):
put in parantheses: negative four to the negative second, the result of the parantheses to the negative one
Let me tell u what I think:
The exponent negative two in the parantheses doesn't affect the negative sign of the four, since the even exponent doesn't affect the negative sign when there are no parantheses
example:
{ (-a)^2 would not be the same as -a^2 }
so the result of the parantheses is -(1/16) and this to the -1 gives you -16
my question (which is probably kind of weird) is (if my explanation is right), if there is a mathematical proof that -a^2 is not the same as (-a)^2 (and that the answer to my problem is -16 and not 16 because of this reason)
I only know that rule, that it doesn't affect the negative sign without parantheses.
I hope I could express myself better this time,
thank u a lot again!!!
Jan!!! |
Jan
| | Veröffentlicht am Mittwoch, den 02. Februar, 2000 - 03:10: |
|
Hey,
first of all, thanks for your effort
I'm sorry to disturb you again,
I mean:
(-4^-2)^-1
(say):
put in parantheses: negative four to the negative second, the result of the parantheses to the negative one
Let me tell u what I think:
The exponent negative two in the parantheses doesn't affect the negative sign of the four, since the even exponent doesn't affect the negative sign when there are no parantheses
example:
{ (-a)^2 would not be the same as -a^2 }
so the result of the parantheses is -(1/16) and this to the -1 gives you -16
my question (which is probably kind of weird) is (if my explanation is right), if there is a mathematical proof that -a^2 is not the same as (-a)^2 (and that the answer to my problem is -16 and not 16 (or anything else) because of this reason)
I only know that rule, that it doesn't affect the negative sign without parantheses.
I hope I could express myself better this time,
thank u a lot again!!!
Jan!!! |
Jan
| | Veröffentlicht am Mittwoch, den 02. Februar, 2000 - 03:17: |
|
Hey,
excuse me that I accidentally sent the same mail three times (I had to make some changes but didn't notice that I had sent it off already, the last one is the final version)
sorry,
Jan!!! |
Fern
| | Veröffentlicht am Mittwoch, den 02. Februar, 2000 - 17:13: |
|
Hello Jan once again,
I'm not sure if I clearly understand what your problem is.
It looks like you mean by -4^2 the negative of (4 squared).
It is best to write this as: -(4^2)
Then: -(4^2)=-(4*4)=-16
However, if we consider (-4)^2 we have
(-4)*(-4)= 4
===================
The exponent "-1" just means to take the inverse of the base number.
a^-1=1/a
and
(-a)^-1=1/(-a)
and
-(a^-1)=-1/a which is the same as 1/(-a)
========================
So, your resolving the first parantheses is correct:
-(4^-2)=-1/16
But your conclusion that this to the power of (-1)
would be 16 is NOT correct.
The inverse number of -1/16 is 1/[-1/16]=-16
==============================================
Hope this can help you. |
Zaph
| | Veröffentlicht am Mittwoch, den 02. Februar, 2000 - 20:58: |
|
Hi Jan,
I guess this is more a problem of the mathematical spelling/convention/agreement/notation than a problem of a missing mathematical proof.
If you write '-a²' then this means (-1)*(a²) = (-1)*(a*a). And if you write '(-a)²' then this means (-a)*(-a). (At this point indeed you can prove (-a)² = (-a)*(-a) = a*a = a², and of course, this is different to (-1)*a² if a² is not equal to zero.)
The problem is similar to 'a+b*c'. This spelling means in the whole mathematical world: first multiply b by c and than add the result to a (not: add a with b and multiply the result by c, which would be denoted by '(a+b)*c'.). And usually you will have a+b*c is NOT equal to (a+b)*c. |
Jan
| | Veröffentlicht am Mittwoch, den 02. Februar, 2000 - 22:03: |
|
Hey,
hey Fern, hey Zaph,
I truly wanted to thank u again, the content of your mails was exactly what I needed,
I anyway think this site and their supporters are doing a great job,
keep it up,
thanks again,
Jan!!!
PS: I meant the "mathematical spelling" Zaph, thanks!!! |
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