anonymous
  • anonymous
The variables x and y satisfy the equation (x)^ny=C, where n and C are constants. When x=1.10, y=5.20, and when x=3.20, y=1.05. (i) Find the values of n and C.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[x ^{n}y=C\]
anonymous
  • anonymous
Michele_Laino
  • Michele_Laino
if we substitute your data, we get two different conditions, namely: \[\Large \begin{gathered} {\left( {1.1} \right)^n} \cdot 5.2 = C \hfill \\ \hfill \\ {\left( {3.2} \right)^n} \cdot 1.05 = C \hfill \\ \end{gathered} \] those equation are an algebraic system, which can be solved for n and C

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Michele_Laino
  • Michele_Laino
equations*
anonymous
  • anonymous
not being able to solve -_-
Michele_Laino
  • Michele_Laino
if I use the elimination method, I can write this: \[\Large {\left( {1.1} \right)^n} \cdot 5.2 = {\left( {3.2} \right)^n} \cdot 1.05\]
anonymous
  • anonymous
yesss
Michele_Laino
  • Michele_Laino
now, I divide both sides of that equation by (1.1)^n, so I get: \[\Large 5.2 = {\left( {\frac{{3.2}}{{1.1}}} \right)^n} \cdot 1.05\]
Michele_Laino
  • Michele_Laino
then I divide both sides again by 1.05, so I can write this: \[\Large \frac{{5.2}}{{1.05}} = {\left( {\frac{{3.2}}{{1.1}}} \right)^n}\]
anonymous
  • anonymous
okay
Michele_Laino
  • Michele_Laino
we got an exponential equation, which can be solved using logarithms
anonymous
  • anonymous
not getting the right answer
Michele_Laino
  • Michele_Laino
why?
anonymous
  • anonymous
i dont know.. can you continue further ?
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • Michele_Laino
if we take the logarithm in base 10, of both sides, we get: \[\Large \begin{gathered} n \cdot {\log _{10}}\left( {\frac{{3.2}}{{1.1}}} \right) = {\log _{10}}\left( {\frac{{5.2}}{{1.05}}} \right) \hfill \\ \hfill \\ n = \frac{{{{\log }_{10}}\left( {\frac{{5.2}}{{1.05}}} \right)}}{{{{\log }_{10}}\left( {\frac{{3.2}}{{1.1}}} \right)}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
what do you get?
anonymous
  • anonymous
finally got the answer.. thank you very much
Michele_Laino
  • Michele_Laino
thanks! :)

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