Mit Logarithmus Gleichung lösen (1/2)2x-3 = (2/3)3x+2

Question

(1/2)^{2x-3} = (2/3)^{3x+2}

wie gehe ich hier vor ?

Answer 1

(1/2) ^{2 x - 3} = ( 2/3 ) ^{3 x + 2}

<=> log ( (1/2) ^{2 x - 3} ) = log ( ( 2/3 ) ^{3 x + 2} )

<=> ( 2 x - 3 ) * log ( 1/2) = ( 3 x + 2 ) * log ( 2/3 )

<=> 2 x * log ( 1/2 ) - 3 * log ( 1/2 ) = 3 x * log ( 2/3 ) + 2 * log ( 2/3)

<=> - 3 * log ( 1/2 ) - 2 * log ( 2/3 ) = 3 x * log ( 2/3 ) - 2 x * log ( 1/2 )

<=> - 3 * log ( 1/2 ) - 2 * log ( 2/3 ) = x * ( 3 * log ( 2/3 ) - 2 * log ( 1/2 ) )

<=> x = ( - 3 * log ( 1/2 ) - 2 * log ( 2/3 ) ) / ( 3 * log ( 2/3 ) - 2 * log ( 1/2 ) )

<=> x = 17,01229...

Answer 2

hi

\(
\left(\frac{1}{2}  \right)^{2x-3} = \left(\frac{2}{3}  \right)^{3x+2} \\
\left(\frac{1}{2}  \right)^{2x}\cdot\left(\frac{1}{2}  \right)^{-3} = \left(\frac{2}{3}  \right)^{3x}\cdot\left(\frac{2}{3}  \right)^{2} \\
\left(\frac{1}{2}  \right)^{2x}\cdot\left(\frac{2}{3}  \right)^{-3x} = \left(\frac{2}{3}  \right)^{2}\cdot \left(\frac{1}{2}  \right)^3 \\
\left(\frac{1}{2}  \right)^{2x}\cdot\left(\frac{2}{3}  \right)^{-3x} = \frac{1}{18} \\
\log { \left( \left(\frac{1}{2}  \right)^{2x}\cdot\left(\frac{2}{3}  \right)^{-3x}  \right) } = \log \frac{1}{18} \\
2x \log \frac{1}{2} - 3x \log \frac{2}{3} = \log \frac{1}{18}\\
x\left( 2 \log \frac{1}{2} - 3 \log \frac{2}{3}  \right)=  \log \frac{1}{18} \\
x\left( \log \left( \left(\frac{1}{2}  \right)^2  \right) - \log \left(\left(\frac{2}{3}  \right)^3  \right)  \right)= \log \frac{1}{18} \\
x \left(\log \frac{1}{4} - \log \frac{8}{27}   \right)= \log \frac{1}{18} \\
x \left(\log \frac{\frac{1}{4}}{\frac{8}{27}} \right) = \log \frac{1}{18} \\
x \left(\log \frac{27}{32} \right)= \log \frac{1}{18} \\
x =   \frac{\log \frac{1}{18} }{\log \frac{27}{32} } \\
x \approx 17,012
\)

Comments

Was mache ich falsch?

Wenn ich x = log (1/18)  /  log (27/32) rechne, bekomme ich immer 12,239277 als Ergebis???

Wo liegt der Fehler?

Wx=  log1 18   log27 32  x=  log1 18   log27 32  x=  log1 18   log27 32 x=  log1 18   log27 32 

---

Hallo Kodakus,

 

Klammern nicht vergessen! (Es macht übrigens keinen Unterschied, ob man log oder ln nimmt)

 

Besten Gruß

Answer 3

$(\frac{1}{2})^{2x-3}= (\frac{2}{3})^{3x+2}$

$(2x-3)\ln(\frac{1}{2})=(3x+2)\ln(\frac{2}{3})$

$2x\cdot \ln(\frac{1}{2})-3\cdot \ln(\frac{1}{2})=3x\cdot \ln(\frac{2}{3}) +2\cdot \ln(\frac{2}{3})$

$- 2x\cdot \ln(2)+3\cdot \ln(2)=3x\cdot \ln(\frac{2}{3}) +2\cdot \ln(\frac{2}{3})$

$3x\cdot \ln(\frac{2}{3}) +2x\cdot \ln(2)= 3\cdot \ln(2)-2\cdot \ln(\frac{2}{3})$

$x\cdot(3 \ln(\frac{2}{3}) +2\cdot \ln(2))= 3\cdot \ln(2)-2\cdot \ln(\frac{2}{3})$

$x=\frac{ 3\cdot \ln(2)-2\cdot \ln(\frac{2}{3})}{3 \ln(\frac{2}{3}) +2\cdot \ln(2)}$

Comments

(1/2)^(2x-3) = 2^(3-2x)

2^(3-2x) = (2/3)^(3x+2)| ln

(3-2x)*ln2 = (3x+2)*ln(2/3)

(3-2x)/(3x+2) =  ln(2/3)/ln2 = z (= -0,5849625)

3-2x = z*(3x+2)

3-2x = 3xz+2z

3xz+2x = 3-2z

x(3z+2) = 3-2z

x = (3-2z)/(3z+2)

x= 17,01