heureka

3.2=1.05^t  what does t equal?

$$\begin{array}{rcl}
3.2 &=& 1.05^{\textcolor[rgb]{1,0,0}{t}} \quad | \quad \ln{}\\ \\
\ln{(3.2)} &=& \ln{(1.05^{\textcolor[rgb]{1,0,0}{t}}} )} \\ \\
\ln{(3.2)} &=& \textcolor[rgb]{1,0,0}{t}} \times \ln{(1.05)} \\ \\
\textcolor[rgb]{1,0,0}{t}} &= &{\ln{(3.2)} \over \ln{(1.05)}
} \\ \\
\textcolor[rgb]{1,0,0}{t}} &= &{1.16315080981\over 0.04879016417
} \\ \\
\textcolor[rgb]{1,0,0}{t}} &= & 23.8398625954
\end{array}$$

e  ^x = 3   ?

$$\begin{array}{rcl}
\\e^x &=& 3 \quad | \quad \ln{} \\
\ln{(e^x)} &=& ln{(3)}\\
x\ln{(e)} &=& ln{(3)} \quad | \quad \ln{(e)} = 1 \;! \\
x &=& ln{(3)}\\
x &=& 1.09861228867
\end{array}$$

$$\boxed{x=\ln{(3)}}$$

$$\\{5\over 8}{8\over5}{4\over 7}{7\over4}{11\over 3}x=1 \\\\
{11\over 3}x=1 \\ \\
x={3\over 11}$$

(1+i)(c+di)=1

$$\begin{array}{lr}
\mbox{Ausklammern:} & c + di + ic + dii = 1\\
\mbox{Ersetzen:} & ii=i^2 = -1 \\
& c + di + ic - d = 1\\
\text{Zusammenfassen:} & \underbrace{ (c-d)}_{=1}+\underbrace{(c+d)}_{=0}i= 1
\end{array}$$

Wenn 1 herauskommen soll, so muss (c  - d) = 1 und (c + d) = 0 sein!

$$\\c + d = 0 \\
\Rightarrow c = - d \\ \\
c - d = 1 \quad | \quad c=-d \quad \mbox{ einsetzen}\\
- d - d = 1 \\
-2d = 1 \\\\
\boxed{ d=-{1\over 2} }$$

$$\\c=-d =-(-{1\over 2})\\\\
\boxed{c={1 \over 2}}$$

$$\\(1+i)(c+di)=1\\ \\
\boxed{(1+i)\left({1 \over 2} -{1 \over 2}i\right) = 1 }$$

writing sinh inverse (-(48/55)) in logarithim form

$$\\\boxed{\sinh^{-1}{(x)}=\ln{(x+ \sqrt{x^2+1} }) }\\ \\
\sinh^{-1}{( -{48 \over 55} )}
= \ln{ \left(-{48 \over 55}
+\sqrt{ (-{48 \over 55} )^2+1}
\right)} \\\\
=\ln{(0.45454545455)}\\\\
=-0.78845736036$$

if r and s are the roots of 2x^2 -5x -7 =0 what is the value of r/s + s/r ?

$$ax^2+bx+c=0 \\
\mbox{The roots are: } x_{1,2}= {-b\pm\sqrt{b^2-4ac}\over2a}\\
\mbox{or } x_{1,2}= {-b\pm\sqrt{D}\over2a} \quad \mbox{ set }\quad D=b^2-4ac\\
\mbox{So we have: } r=x_1={-b+\sqrt{D}\over 2a}\\
\mbox{and } s=x_2={-b-\sqrt{D}\over 2a}$$

$$\\ \boxed{{r\over s}+{s\over r} =} {-b+\sqrt{D}\over -b-\sqrt{D}}+{-b-\sqrt{D}\over -b+\sqrt{D}}\\ \\
={(-b+\sqrt{D})^2 +(-b-\sqrt{D})^2\over (-b-\sqrt{D})(-b+\sqrt{D})}\\\\
={2(-b)^2+2(\sqrt{D})^2 \over (-b)^2-(\sqrt{D})^2}\\\\
={2b^2+2D \over b^2-D} \quad | \quad D= b^2-4ac$$

$$\\={2b^2+2b^2-8ac \over b^2-b^2+4ac} \\\\
={4b^2-8ac \over 4ac} \\\\
={b^2 \over ac} - 2 \quad | \quad a=2 \quad b=-5 \quad c=-7$$

$$\\={ (-5)^2 \over 2*(-7) } -2 \\\\
={ 25 \over (-14) } -2 \\\\
={ -25-2*14 \over 14 } \\\\
={ -\frac{53} {14} } \\\\
={ -\frac{14*3+11} {14} } \\\\
={ -3-\frac{11} {14} } \\\\
\boxed{={ -3\frac{11} {14} }}$$

1. Number: i

2. Number: i+1

3. Number: i+2

$$\\{(2.\;Number + 3.\;Number) \over 2 }= 35\\\\
{((i+1)+ (i+2)) \over 2 }= 35 \\\\
{(2i+1+2)\over 2 }= 35 \\\\
{(2i+1) \over 2 }+1 = 35 \\\\
{(2i+1) \over 2 } =34 \quad | \quad \mbox{2i+1 is an odd number}$$

an odd number cannot be divided by 2

Here is no solution!

0,63 stunden sind wieviele minuten?

$$\\0,63\;Stunden * {60\;Minuten\over 1\; Stunde} \quad | \quad \mbox{60 Minuten sind eine Stunde}\\\\
=0.63 * 60\;Minuten
= 37.8\; Minuten\\\\
\textcolor[rgb]{1,0,0}{\mbox{0,63 Stunden sind 37.8 Minuten}}$$

$$\boxed{{\ d{(\ln{y})} \over \ dx }={y' \over y }}\\\\
\Rightarrow {\ d{(\ln{x})} \over \ dx }={1 \over x }\\\\
\Rightarrow {\ d({\ln{(a\times x)}}) \over \ dx } ={a \over a\times x } ={1 \over x }$$

$$\begin{array}{ccl}
\int{ {1\over x} \ dx }&=& \ln{(|a\times x|)} +c_1\\
& = &\ln{(|a|)}+\ln{(|x|)}+c_1 \\
& = &\ln{(|x|)}+\underbrace{\ln{(|a|)}+c_1}_{c_2} \\
& = &\ln{(|x|)}+c_2}
\end{array}$$

distance between points (2,-5) and (-3,0) ?

$$\begin{array}{crcrccr}
\mbox{Point 1:} \; &(x_1&=&-3,& y_1&=&0)\\
\mbox{Point 2:} \; &(x_2&=&2,& y_2&=&-5)
\end{array}$$

$$\begin{array}{ccl}
distance&=&\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\
distance&=&\sqrt{[2-(-3)]^2+[(-5)-0]^2}\\
&=&\sqrt{(2+3)^2+(-5)^2}\\
&=&\sqrt{(5)^2+(-5)^2}\\
&=&\sqrt{25+25}\\
&=&\sqrt{2*25}\\
&=&\sqrt{25*2}\\
&=&\sqrt{25}\sqrt{2}\\
distance&=&5\sqrt{2}\\
\end{array}$$