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geerky42

  • one year ago

Find the value of x and y:\[ (2x)^{\ln 2} = (3y)^{\ln 3} \]\[ 3^{\ln x} = 2^{\ln y}\]

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  1. Goten77
    • one year ago
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    hmm

  2. Goten77
    • one year ago
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    |dw:1358145356088:dw|

  3. wio
    • one year ago
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    Is this a system of equations?

  4. Goten77
    • one year ago
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    |dw:1358145846151:dw|

  5. Goten77
    • one year ago
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    |dw:1358147158729:dw|

  6. wio
    • one year ago
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    .... @Goten77 hard to read it... what are you doing?

  7. Goten77
    • one year ago
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    |dw:1358147202405:dw|

  8. Goten77
    • one year ago
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    think bout it..... not that hard to read..... but it wont ever end....

  9. wio
    • one year ago
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    \[ \large \begin{array}{rcl} (2x)^{\ln2}&=&(3y)^{\ln3} \\ \ln(2x)\ln(2)&=&\ln(3y)\ln(3) \\ \ln(2)\ln(2) +\ln(2) \ln(x) &=&\ln(3)\ln(y)+\ln(3)\ln(3) \\ \ln(2) \ln(x) - \ln(3)\ln(y) &=&\ln(3)\ln(3)-\ln(2)\ln(2) \end{array} \] \[ \large \begin{array}{rcl} 3^{\ln x}&=&2^{\ln y} \\ \ln(3)\ln (x)&=&\ln(2)\ln (y) \\ \ln(3)\ln (x) - \ln(2)\ln (y) &=& 0 \end{array} \] \[ \begin{bmatrix} \ln(2)&-\ln(3) \\ \ln(3)&- \ln(2) \end{bmatrix} \begin{bmatrix} \ln(x) \\ \ln(y) \end{bmatrix} = \begin{bmatrix} \ln(3)\ln(3)-\ln(2)\ln(2) \\ 0 \end{bmatrix} \]

  10. wio
    • one year ago
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    If you put it into a matrix, it's not as big a mess.

  11. wio
    • one year ago
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    then at the very end raise it to the power of \(e\). It could be a singular matrix though?

  12. Goten77
    • one year ago
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    strange its basically a property..... memorable

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