heureka

Three consecutive integers have a sum of 90. Find the three values?

\(\begin{array}{rcll} (n-1) + n + ( n+1) &=& 90 \\ 3n &=& 90 \\ n &=& 30 \\\\ n_1 = n-1 = 30 - 1 &=& 29 \\ n_2 = n &=& 30 \\ n_3 = n+1 = 30 + 1 &=& 31 \\ \end{array}\)

x^3-6x=6  ...   x=2^(2/3)+2^(1/3)

\(\boxed{~ \begin{array}{rcll} x^3 + px + q &=& 0 \\ x &=& \sqrt[3]{ -\frac{q}{2} + \sqrt{ \frac{q^2}{4} + \frac{p^3}{27} } } + \sqrt[3]{ -\frac{q}{2} - \sqrt{ \frac{q^2}{4} + \frac{p^3}{27} } } \\ \text{If } \frac{q^2}{4} + \frac{p^3}{27} \ge 0 \end{array} ~}\)

\(\begin{array}{rcll} x^3-6x &=& 6 \\ x^3-6x -6 &=& 0 \\\\ x^3 + px + q &=& 0 \quad | \quad p = -6 = \quad q = -6 \\ x &=& \sqrt[3]{ -\frac{q}{2} + \sqrt{ \frac{q^2}{4} + \frac{p^3}{27} } } + \sqrt[3]{ -\frac{q}{2} - \sqrt{ \frac{q^2}{4} + \frac{p^3}{27} } } \\ x &=& \sqrt[3]{ -\frac{-6}{2} + \sqrt{ \frac{(-6)^2}{4} + \frac{(-6)^3}{27} } } + \sqrt[3]{ -\frac{-6}{2} - \sqrt{ \frac{(-6)^2}{4} + \frac{(-6)^3}{27} } } \\ x &=& \sqrt[3]{ 3 + \sqrt{ 9 - \frac{(6)^3}{27} } } + \sqrt[3]{ 3 - \sqrt{ 9 - \frac{(6)^3}{27} } } \\ x &=& \sqrt[3]{ 3 + \sqrt{ 9 - \frac{8\cdot 27}{27} } } + \sqrt[3]{ 3 - \sqrt{ 9 - \frac{8\cdot 27}{27} } } \\ x &=& \sqrt[3]{ 3 + \sqrt{ 9 - 8 } } + \sqrt[3]{ 3 - \sqrt{ 9 - 8 } } \\ x &=& \sqrt[3]{ 3 + \sqrt{ 1 } } + \sqrt[3]{ 3 - \sqrt{ 1 } } \\ x &=& \sqrt[3]{ 3 + 1 } + \sqrt[3]{ 3 - 1 } \\ x &=& \sqrt[3]{ 4 } + \sqrt[3]{ 2 } \\ x &=& \sqrt[3]{ 2^2 } + \sqrt[3]{ 2 } \\ x &=& 2^{\frac23} + 2^{\frac13} \\ \end{array}\)

3^(2x)+9=10*3^x

\(\begin{array}{rcll} 3^{2x}+9 &=& 10\cdot 3^x \qquad & | \qquad - 10\cdot 3^x \\ 3^{2x}- 10\cdot 3^x + 9 &=& 0 \\ 3^{x+x}- 10\cdot 3^x + 9 &=& 0 \\ 3^x \cdot 3^x - 10\cdot 3^x + 9 &=& 0 \qquad & | \qquad z= 3^x\\ z\cdot z - 10\cdot z + 9 &=& 0 \\ z^2 - 10\cdot z + 9 &=& 0 \\ \end{array}\)

\( \boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0 \\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~} \)

\(\begin{array}{rcll} z^2 - 10\cdot z + 9 &=& 0 \quad | \quad a=1 \quad b = -10 \quad c = 9 \\ z &=& \dfrac{-(-10) \pm \sqrt{(-10)^2-4\cdot 1 \cdot 9} }{2\cdot 1} \\ z &=& \dfrac{ 10 \pm \sqrt{100-36} }{2} \\ z &=& \dfrac{ 10 \pm \sqrt{64} }{2} \\ z &=& \dfrac{ 10 \pm 8 }{2} \\ z &=& 5 \pm 4 \\\\ z_1 &=& 5+4 \\ \mathbf{z_1} &\mathbf{=}& \mathbf{9} \\\\ z_2 &=& 5-4 \\ \mathbf{z_2} &\mathbf{=}& \mathbf{1} \end{array}\)

\(\begin{array}{rcll} z &=& 3^x \qquad & | \qquad \ln{()} \\ \ln{(z)} &=& \ln{(3^x)} \\ \ln{(z)} &=& x \cdot \ln{(3)} \\ x &=& \frac{ \ln{(z)} }{ \ln{(3)} } \\\\ x_1 &=& \frac{ \ln{(z_1)} }{ \ln{(3)} } \\ x_1 &=& \frac{ \ln{(9)} }{ \ln{(3)} } \\ x_1 &=& \frac{ \ln{(3^2)} }{ \ln{(3)} } \\ x_1 &=& \frac{ 2 \ln{(3)} }{ \ln{(3)} } \\ \mathbf{x_1} &\mathbf{=}& \mathbf{2} \\\\ x_2 &=& \frac{ \ln{(z_2)} }{ \ln{(3)} } \\ x_2 &=& \frac{ \ln{(1)} }{ \ln{(3)} } \qquad & | \qquad \ln{(1)} = 0\\ x_2 &=& \frac{ 0 }{ \ln{(3)} } \\ \mathbf{x_2} &\mathbf{=}& \mathbf{0} \\\\ \end{array}\)

6(4^x+9^x)=13*6^x

\(\begin{array}{rcll} 6\cdot (4^x+9^x) &=& 13\cdot 6^x \qquad & | \qquad :6\\ 4^x+9^x &=& \frac{13}{6}\cdot 6^x \qquad & | \qquad :6^x\\ \frac{ 4^x } {6^x} + \frac{ 9^x } {6^x} &=& \frac{13}{6}\\ \left(\frac46 \right)^x + \left(\frac96 \right)^x &=& \frac{13}{6}\\ \left(\frac23 \right)^x + \left(\frac32 \right)^x &=& \frac{13}{6}\\ \left( \frac{1}{\frac32} \right)^x + \left(\frac32 \right)^x &=& \frac{13}{6}\\ \frac{1}{ \left( \frac32\right)^x } + \left(\frac32 \right)^x &=& \frac{13}{6} \qquad & | \qquad z= \left(\frac32 \right)^x\\ \frac{1}{ z } + z &=& \frac{13}{6} \qquad & | \qquad \cdot z\\ \frac{z}{ z } + z^2 &=& \frac{13}{6} \cdot z\\ 1 + z^2 &=& \frac{13}{6} \cdot z\\ z^2 - \frac{13}{6} \cdot z + 1 &=& 0 \\ \end{array}\)

\(\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0 \\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~}\)

\(\begin{array}{rcll} z^2 - \frac{13}{6} \cdot z + 1 &=& 0 \quad | \quad a=1 \quad b =-\frac{13}{6} \quad c = 1 \\ z &=& \dfrac{-(-\frac{13}{6}) \pm \sqrt{(-\frac{13}{6})^2-4\cdot 1 \cdot 1} }{2\cdot 1} \\ z &=& \dfrac{ \frac{13}{6} \pm \sqrt{ \frac{13^2}{6^2}-4 } }{2} \\ z &=& \dfrac{ \frac{13}{6} \pm \sqrt{ \frac{13^2-4\cdot 6^2}{6^2} } }{2} \\ z &=& \dfrac{ \frac{13}{6} \pm \sqrt{ \frac{169-144}{6^2} } }{2} \\ z &=& \dfrac{ \frac{13}{6} \pm \sqrt{ \frac{25}{6^2} } }{2} \\ z &=& \dfrac{ \frac{13}{6} \pm \sqrt{ \frac{5^2}{6^2} } }{2} \\ z &=& \dfrac{ \frac{13}{6} \pm \frac{5}{6} }{2} \\ z &=& \frac{13}{12} \pm \frac{5}{12} \\ z_1 &=& \frac{13}{12} + \frac{5}{12} \\ z_1 &=& \frac{18}{12} \\ \mathbf{z_1} &\mathbf{=}& \mathbf{\frac{3}{2}} \\\\ z_2 &=& \frac{13}{12} - \frac{5}{12} \\ z_2 &=& \frac{8}{12} \\ \mathbf{z_2} &\mathbf{=}& \mathbf{\frac{2}{3}} \end{array}\)

\(\begin{array}{rcll} z&=& \left(\frac32 \right)^x \qquad & | \qquad \ln{()} \\ \ln{(z)} &=& x \cdot \ln{ \left(\frac32 \right) } \\ x &=& \frac{ \ln{(z)} }{ \ln{ \left(\frac32 \right) } } \\\\ x_1 &=& \frac{ \ln{(z_1)} }{ \ln{ \left(\frac32 \right) } } \\ x_1 &=& \frac{ \ln{\left(\frac32 \right)} }{ \ln{ \left(\frac32 \right) } } \\ \mathbf{x_1} &\mathbf{=}& \mathbf{1} \\\\ x_2 &=& \frac{ \ln{(z_2)} }{\ln{ \left(\frac32 \right) } } \\ x_2 &=& \frac{ \ln{ \left(\frac23 \right) } }{ \ln{ \left(\frac32 \right) } } \\ x_2 &=& \frac{ \ln{ \left( \frac{1}{ \frac32} \right) } }{ \ln{ \left(\frac32 \right) } } \\ x_2 &=& \frac{ \ln{(1)} - \ln{ \left( \frac32 \right) } }{ \ln{ \left(\frac32 \right) } } \qquad & | \qquad \ln{(1)} = 0\\ x_2 &=& \frac{ 0 - \ln{ \left( \frac32 \right) } }{ \ln{ \left(\frac32 \right) } } \\ x_2 &=& \frac{ -\ln{ \left( \frac32 \right) } }{ \ln{ \left(\frac32 \right) } } \\ \mathbf{x_2} &\mathbf{=}& \mathbf{-1} \\\\ \end{array}\)

4^x-3*2^(x+2)+2^5=0

\(\begin{array}{rcll} 4^x-3\cdot 2^{x+2}+2^5 &=& 0 \\ 4^x-3\cdot 2^x\cdot 2^2 +2^5 &=& 0 \\ 4^x-12\cdot 2^x +2^5 &=& 0 \\ (2\cdot 2)^x-12\cdot 2^x +2^5 &=& 0 \\ 2^x \cdot 2^x-12\cdot 2^x +2^5 &=& 0 \qquad & | \qquad z=2^x\\ z^2 -12z +32 &=& 0 \end{array}\)

\(\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0 \\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~}\)

\(\begin{array}{rcll} z^2 -12z +32 &=& 0 \quad | \quad a=1 \quad b = -12 \quad c = 32 \\ z &=& \dfrac{-(-12) \pm \sqrt{(-12)^2-4\cdot 1 \cdot 32} }{2\cdot 1} \\ z &=& \dfrac{ 12 \pm \sqrt{144-128} }{2} \\ z &=& \dfrac{ 12 \pm \sqrt{^16} }{2} \\ z &=& \dfrac{ 12 \pm 4 }{2} \\ z &=& 6 \pm 2 \\\\ z_1 &=& 6+2 \\ \mathbf{z_1} &\mathbf{=}& \mathbf{8} \\\\ z_2 &=& 6-2 \\ \mathbf{z_2} &\mathbf{=}& \mathbf{4} \end{array}\)

\(\begin{array}{rcll} z &=& 2^x \qquad & | \qquad \ln{()} \\ \ln{(z)} &=& x \ln{(2)} \\ \ln{(z)} &=& x \ln{(2)} \\ x &=& \frac{ \ln{(z)} }{ \ln{(2)} } \\\\ x_1 &=& \frac{ \ln{(z_1)} }{ \ln{(2)} } \\ x_1 &=& \frac{ \ln{(8)} }{ \ln{(2)} } \\ x_1 &=& \frac{ \ln{(2^3)} }{ \ln{(2)} } \\ x_1 &=& \frac{ 3 \ln{(2)} }{ \ln{(2)} } \\ \mathbf{x_1} &\mathbf{=}& \mathbf{3} \\\\ x_2 &=& \frac{ \ln{(z_2)} }{ \ln{(2)} } \\ x_2 &=& \frac{ \ln{(4)} }{ \ln{(2)} } \\ x_2 &=& \frac{ \ln{(2^2)} }{ \ln{(2)} } \\ x_2 &=& \frac{ 2 \ln{(2)} }{ \ln{(2)} } \\ \mathbf{x_2} &\mathbf{=}& \mathbf{2} \\\\ \end{array}\)

a^x=x^y

a^y=x^x

\(\begin{array}{rcll} (1) & a^x &=& x^y \qquad & | \qquad \ln{()} \\ & x\cdot \ln{(a)} &=& y \cdot \ln{(x)} \\ & y &=& \frac{x\cdot \ln{(a)} }{ \ln{(x)} } \\ \hline \\ (2) & a^y &=& x^x \qquad & | \qquad \ln{()} \\ & y\cdot \ln{(a)} &=& x \cdot \ln{(x)} \qquad & | \qquad y = \frac{x\cdot \ln{(a)} }{ \ln{(x)} }\\\\ & \frac{x\cdot \ln{(a)} }{ \ln{(x)} }\cdot \ln{(a)} &=& x \cdot \ln{(x)} \\ & x\cdot [\ln{(a)}]^2 &=& x \cdot [\ln{(x)}]^2 \\ & x\cdot [\ln{(a)}]^2 - x \cdot [\ln{(x)}]^2 &=& 0 \\ & \underbrace{x}_{=0} \cdot \underbrace{ \{~ [\ln{(a)}]^2 - [\ln{(x)}]^2 ~\} }_{=0} &=& 0 \\\\ & \mathbf{x_1} & \mathbf{=} & \mathbf{0} \\\\ & \{~ [\ln{(a)}]^2 - [\ln{(x)}]^2 ~\} &=& 0 \\ & [\ln{(a)}]^2 - [\ln{(x)}]^2 &=& 0 \\ & [\ln{(a)}]^2 &=& [\ln{(x)}]^2 \\ & \ln{(a)} &=& \ln{(x)} \\ & a &=& x \\ & \mathbf{x_2} & \mathbf{=} & \mathbf{a} \\ \hline \\ & y &=& \frac{x\cdot \ln{(a)} }{ \ln{(x)} } \\ & y &=& \frac{a\cdot \ln{(a)} }{ \ln{(a)} } \\ & \mathbf{y} &\mathbf{=}& \mathbf{a} \\ \end{array}\)

-11 modulo 10 = ¿?

\(\boxed{~ \begin{array}{rcll} a \pmod b &=& a - b \cdot \left[ \frac{a}{b} \right] \qquad [ \dots ] = \text{ integer part} \end{array} ~}\)

\(\begin{array}{rcll} a \pmod b &=& a - b \cdot \left[ \frac{a}{b} \right] \qquad & | \qquad a = (-11) \qquad b = 10 \\\\ (-11) \pmod {10} &=& (-11) - 10 \cdot \left[ \frac{-11}{10} \right] \\ &\equiv& (-11) - 10 \cdot [ -1.1 ] \\ &\equiv& (-11) - 10 \cdot ( -1) \\ &\equiv& (-11) + 10 \\ &\equiv& -1 \\ &\equiv& -1 + 10 \\ &\equiv& 9 \\ (-11) \pmod {10} &=& 9 \end{array}\)

log(384/5)+log(81/32)+3log(5/3)+log(1/9)=?

\(\begin{array}{rcll} \log{ \left( \frac{384}{5} \right) } + \log{ \left( \frac{81}{32} \right) } +3 \cdot \log{ \left( \frac{5}{3} \right) } + \log{ \left( \frac{1}{9} \right) } &=&\\ &=& \log{ \left( \frac {384}{5}\cdot \frac {81}{32} \cdot \frac {5^3}{3^3} \cdot \frac{1}{9} \right) } \\ &=& \log{ \left( \frac {3\cdot 2^7}{5}\cdot \frac {3^4}{2^5} \cdot \frac {5^3}{3^3} \cdot \frac{1}{3^2} \right) } \\ &=& \log{ \left( \frac{2^7}{2^5} \cdot \frac{3\cdot 3^4 }{3^3\cdot 3^2} \cdot \frac{5^3}{5} \right) } \\ &=& \log{ \left( \frac{2^7}{2^5} \cdot \frac{ 3^5 }{3^5} \cdot \frac{5^3}{5} \right) } \\ &=& \log{ \left( \frac{2^7}{2^5} \cdot \frac{5^3}{5} \right) } \\ &=& \log{ ( 2^2 \cdot 5^2 ) } \\ &=& \log{ ( (2\cdot 5 )^2 ) } \\ &=& \log{ ( 10^2 ) } \\ &=& 2\cdot \log{ ( 10 ) } \qquad & | \qquad \log{ ( 10 ) } = 1\\ &=& 2\cdot 1\\ \log{ \left( \frac{384}{5} \right) } + \log{ \left( \frac{81}{32} \right) } +3 \cdot \log{ \left( \frac{5}{3} \right) } + \log{ \left( \frac{1}{9} \right) } &=& 2 \end{array}\)

ln ist der natürliche Logarithmus zur Basis e.

\(\begin{array}{rcll} x &=& - \frac{ \ln{ \left(\frac54 \right) } }{ \ln{ \left(\frac{\sqrt{5}}{2} - \frac12 \right) } } \\\\ &=& - \frac{ \ln{ \left(\frac54 \right) } }{ \ln{ \left( \frac12 \cdot ( \sqrt{5}-1 ) \right) } } \\\\ &=& - \frac{ \ln{ (1,25 ) } }{ \ln{ \left( \frac12 \cdot 1,23606797750 \right) } } \\\\ &=& - \frac{ \ln{ (1,25 ) } }{ \ln{( 0,61803398875 ) } } \\\\ &=& - \frac{ 0.22314355131 }{ -0.48121182506 } \\\\ &=& \frac{ 0.22314355131 }{ 0.48121182506 } \\\\ x &=& 0.46371169554 \end{array}\)

cos2x - cosx = 0 i need steps....

rearrange

\(\begin{array}{rcll} \cos{(2x)} - \cos{(x)} &=& 0 \\ \cos{(2x)} &=& \cos{(x)} \\\\ \boxed{~ \cos{(\varphi)} = \cos{(-\varphi)}~}\\\\ \end{array}\)

We have 4 variations: {nl} \(\begin{array}{lrclrcl} (1) & \cos{(2x)} &=& \cos{(x)} \qquad \Rightarrow &\qquad 2x \pm 2k_1\pi &=& x \pm 2k_2\pi\\ (2) & \cos{(-2x)} &=& \cos{(x)} \qquad \Rightarrow &\qquad -2x \pm 2k_1\pi &=& x \pm 2k_2\pi\\ (3) & \cos{(2x)} &=& \cos{(-x)} \qquad \Rightarrow &\qquad 2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\\ (4) & \cos{(-2x)} &=& \cos{(-x)} \qquad \Rightarrow &\qquad -2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\\\\ (1) & 2x \pm 2k_1\pi &=& x \pm 2k_2\pi\qquad \Rightarrow &\qquad x &=& 0 \pm 2k\pi\\ (2) & -2x \pm 2k_1\pi &=& x \pm 2k_2\pi \qquad \Rightarrow &\qquad -3x &=& 0 \pm 2k\pi\\ (3) & 2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\qquad \Rightarrow &\qquad 3x &=& 0 \pm 2k\pi\\ (4) & -2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\qquad \Rightarrow &\qquad -x &=& 0 \pm 2k\pi\\\\ & \Rightarrow x &=& \pm 2k\pi \\ & \Rightarrow x &=& \pm \frac{2}{3}k\pi \\ \end{array}\)

We can put together  \(x = \pm \frac{2}{3}k\pi \qquad k \in N_0\)