Find exponential models y_1=Ce^{k_1t} and y_2=C(2)^{k_2t} That pass through the two given points. Compare the values of k_1 and k_2. (If you round your answer, round to four decimal places.) (0,16),(60,frac{1}{4}) y_1=Ce^{k_1t} where C=? and k_1=? y_2=C(2)^{k_2t} where C=? and k_2=?

Question
Exponential models
asked 2021-03-06
Find exponential models
\(\displaystyle{y}_{{1}}={C}{e}^{{{k}_{{1}}{t}}}\)
and
\(\displaystyle{y}_{{2}}={C}{\left({2}\right)}^{{{k}_{{2}}{t}}}\)
That pass through the two given points. Compare the values of \(\displaystyle{k}_{{1}}\) and \(\displaystyle{k}_{{2}}\). (If you round your answer, round to four decimal places.)
\(\displaystyle{\left({0},{16}\right)},{\left({60},{\frac{{{1}}}{{{4}}}}\right)}\)
\(\displaystyle{y}_{{1}}={C}{e}^{{{k}_{{1}}{t}}}\) where C=?
and \(\displaystyle{k}_{{1}}=?\)
\(\displaystyle{y}_{{2}}={C}{\left({2}\right)}^{{{k}_{{2}}{t}}}\) where C=?
and \(\displaystyle{k}_{{2}}=?\)

Answers (1)

2021-03-07
Given equations are \(\displaystyle{y}_{{1}}={C}{e}^{{{k}_{{1}}{t}}}\)
\(\displaystyle{y}_{{2}}={C}{\left({2}^{{{k}_{{2}}{t}}}\right)}\)
They pass through points (0, 16) and \(\displaystyle{\left({60},{\frac{{{1}}}{{{4}}}}\right)}\)
now \(\displaystyle{y}_{{1}}={C}{e}^{{{k}_{{1}}{t}}},{\left({0},{16}\right)}\)
\(\displaystyle{16}={C}{e}^{{0}}={C}\)
\(\displaystyle{C}={16}\)
\(\displaystyle{y}_{{1}}={16}{e}^{{{k}_{{1}}{t}}},{\left({60},{\frac{{{1}}}{{{4}}}}\right)}\)
\(\displaystyle{\frac{{{1}}}{{{4}}}}={16}{e}^{{{60}{k}_{{1}}}}\)
\(\displaystyle{e}^{{{60}{k}_{{1}}}}={\frac{{{1}}}{{{64}}}}\)
\(\displaystyle{60}{k}_{{1}}={\ln{{\frac{{{1}}}{{{64}}}}}}=-{4.158883}\)
\(\displaystyle{k}_{{1}}={\frac{{-{4.158883}}}{{{60}}}}=-{0.0693}\)
\(\displaystyle{k}_{{1}}=-{0.0693}\)
Again
\(\displaystyle{y}_{{1}}={C}{2}^{{{k}_{{2}}{t}}},\)
\(\displaystyle{16}={C}{2}^{{0}}={C}\)
\(\displaystyle{C}={16}\)
\(\displaystyle{y}_{{1}}={16}\times{2}^{{{k}_{{2}}{t}}}\)
\(\displaystyle{\frac{{{1}}}{{{4}}}}={16}\times{2}^{{{60}{k}_{{2}}}}\)
\(\displaystyle{2}^{{{60}{k}_{{2}}}}={\frac{{{1}}}{{{64}}}}={2}^{{-{6}}}\)
\(\displaystyle{60}{k}_{{2}}=-{6}\)
\(\displaystyle{k}_{{2}}=-{\frac{{{6}}}{{{60}}}}=-{0.1}\)
\(\displaystyle{k}_{{2}}=-{0.1}\)
\(\displaystyle{C}={16}\)
\(\displaystyle{k}_{{1}}=-{0.0693}\)
\(\displaystyle{k}_{{2}}=-{0.1}\)
0

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