Edit detail for SandBox ProblemSolving revision 1 of 3

changed:
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  Mathe-Abi ab 2004

Aufgabe: Löse die Gleichung $e^{2x}+e^{x}-2=0$
 

Lösung: 
\begin{axiom}
g:=%e**(2*x)+%e**(x)-2=0
g:=g+2
f:=(2*x^2+2*x-1)/(x^2-1)
differentiate(f,x)
simplify(%)
f2:=3*(x^2-4*%e^(-2*x))
integrate(f2,x)
\end{axiom}

Comparar $ea(n)$ con ordenes como $n^k$, $2^n$, $n!$, $n^n$ donde
$ea(n)=\frac{1}{n-\log_2{2^n-1}}$

Intento:
\begin{axiom}
e:=1/(n-log(2^n-1))
simplify(e)
\end{axiom}

\begin{axiom}
solve(n+2-log((2^(n+1)-1)^2/(2^n-1));n)
\end{axiom}

Mathe-Abi ab 2004

Aufgabe: Löse die Gleichung

Lösung:

axiom
g:=%e**(2*x)+%e**(x)-2=0

(1)

Type: Equation Expression Integer

axiom
g:=g+2

(2)

Type: Equation Expression Integer

axiom
f:=(2*x^2+2*x-1)/(x^2-1)

(3)

Type: Fraction Polynomial Integer

axiom
differentiate(f,x)

(4)

Type: Fraction Polynomial Integer

axiom
simplify(%)

(5)

Type: Expression Integer

axiom
f2:=3*(x^2-4*%e^(-2*x))

(6)

Type: Expression Integer

axiom
integrate(f2,x)

(7)

Type: Union(Expression Integer,...)

Comparar con ordenes como , , , donde

Intento:

axiom
e:=1/(n-log(2^n-1))

(8)

Type: Expression Integer

axiom
simplify(e)

(9)

Type: Expression Integer

axiom
solve(n+2-log((2^(n+1)-1)^2/(2^n-1));n)

(10)

Type: List Equation Fraction Polynomial Integer