logarithmic transformation from exponential to linear equation












0














How to convert this exponential equation to linear equation.



Y =exp(17.9348)*x^(-2.705)



what I did is:
Y =log(17.9348)+(-2.705)*log(x)



I am confused with this one.
Y=17.9348+(-2.705)*log(x)



which one is correct transformation?



Thanks in advance!






logarithms linear-transformations transformation







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    0














    How to convert this exponential equation to linear equation.



    Y =exp(17.9348)*x^(-2.705)



    what I did is:
    Y =log(17.9348)+(-2.705)*log(x)



    I am confused with this one.
    Y=17.9348+(-2.705)*log(x)



    which one is correct transformation?



    Thanks in advance!






    logarithms linear-transformations transformation







    share|cite|improve this question







    New contributor




    user3063 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      0












      0








      0







      How to convert this exponential equation to linear equation.



      Y =exp(17.9348)*x^(-2.705)



      what I did is:
      Y =log(17.9348)+(-2.705)*log(x)



      I am confused with this one.
      Y=17.9348+(-2.705)*log(x)



      which one is correct transformation?



      Thanks in advance!






      logarithms linear-transformations transformation







      share|cite|improve this question







      New contributor




      user3063 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      How to convert this exponential equation to linear equation.



      Y =exp(17.9348)*x^(-2.705)



      what I did is:
      Y =log(17.9348)+(-2.705)*log(x)



      I am confused with this one.
      Y=17.9348+(-2.705)*log(x)



      which one is correct transformation?



      Thanks in advance!






      logarithms linear-transformations transformation




      logarithms linear-transformations transformation






      share|cite|improve this question







      New contributor




      user3063 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question







      New contributor




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      asked 2 days ago









      user3063

      1071




      1071




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          The answer is neither! You need to take the logarithm of both sides:
          begin{align*}
          y&=e^{17.9348},x^{-2.705} \
          ln(y)&=lnleft(e^{17.9348},x^{-2.705}right) \
          ln(y)&=lnleft(e^{17.9348}right)+lnleft(x^{-2.705}right) \
          ln(y)&=17.9348-2.705ln(x).
          end{align*}






          share|cite|improve this answer





















          • ok, thank you so much!
            – user3063
            2 days ago











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          1 Answer
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          The answer is neither! You need to take the logarithm of both sides:
          begin{align*}
          y&=e^{17.9348},x^{-2.705} \
          ln(y)&=lnleft(e^{17.9348},x^{-2.705}right) \
          ln(y)&=lnleft(e^{17.9348}right)+lnleft(x^{-2.705}right) \
          ln(y)&=17.9348-2.705ln(x).
          end{align*}






          share|cite|improve this answer





















          • ok, thank you so much!
            – user3063
            2 days ago
















          1














          The answer is neither! You need to take the logarithm of both sides:
          begin{align*}
          y&=e^{17.9348},x^{-2.705} \
          ln(y)&=lnleft(e^{17.9348},x^{-2.705}right) \
          ln(y)&=lnleft(e^{17.9348}right)+lnleft(x^{-2.705}right) \
          ln(y)&=17.9348-2.705ln(x).
          end{align*}






          share|cite|improve this answer





















          • ok, thank you so much!
            – user3063
            2 days ago














          1












          1








          1






          The answer is neither! You need to take the logarithm of both sides:
          begin{align*}
          y&=e^{17.9348},x^{-2.705} \
          ln(y)&=lnleft(e^{17.9348},x^{-2.705}right) \
          ln(y)&=lnleft(e^{17.9348}right)+lnleft(x^{-2.705}right) \
          ln(y)&=17.9348-2.705ln(x).
          end{align*}






          share|cite|improve this answer












          The answer is neither! You need to take the logarithm of both sides:
          begin{align*}
          y&=e^{17.9348},x^{-2.705} \
          ln(y)&=lnleft(e^{17.9348},x^{-2.705}right) \
          ln(y)&=lnleft(e^{17.9348}right)+lnleft(x^{-2.705}right) \
          ln(y)&=17.9348-2.705ln(x).
          end{align*}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Adrian Keister

          4,79851933




          4,79851933












          • ok, thank you so much!
            – user3063
            2 days ago


















          • ok, thank you so much!
            – user3063
            2 days ago
















          ok, thank you so much!
          – user3063
          2 days ago




          ok, thank you so much!
          – user3063
          2 days ago










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