Find a formula for f(x), an exponential function such that f(-5) = 240 and f(40) = 580.
f(x)=________
+26400 Find a formula for f(x), an exponential function such that f(-5) = 240 and f(40) = 580. f(x)=________
\(\small{ \begin{array}{lrcl} (1) & 580 &=& a\cdot b^{40} \\ (2) & 240 &=& a\cdot b^{-5} \\ \\ \hline \\ (1) : (2) & \frac{580}{240} &=& \frac{ b^{40} }{ b^{-5} } \\ & \frac{580}{240} &=& b^{40}\cdot b^{-5}\\ & \frac{580}{240} &=& b^{45}\\ & \frac{58}{24} &=& b^{45}\\ & \frac{29}{12} &=& b^{45}\\ & b &=& \sqrt[45]{\frac{29}{12}} \\ & b &=& \sqrt[45]{2.416\overline{6}} \\ & b &=& 2.416\overline{6}^{~(0.02\overline{2})}\\ & \mathbf{b} & \mathbf{=} & \mathbf{1.01980216076} \\ \\ \hline \\ & a &=& \frac{580} { b^{40} } \\ & a &=& \frac{580} { 2.19097343941 }\\ & \mathbf{a} & \mathbf{=} & \mathbf{264.722515374} \\ \\ \hline \\ & f(x) &=& a \cdot b^x \\ & f(x) &=& 264.722515374 \cdot 1.01980216076^x \end{array} }\)
\(\small{ \text{A function of the form }\\ f(x) = a\cdot b^x \\ \text{ can be re-written as } \\ f(x) = b^{x ± c} \\ \text{ by the use of logarithms and so is an exponential function.} } \)
\(\small{\text{$ \begin{array}{rcl} 264.722515374 \cdot 1.01980216076^x &=& 1.01980216076^{x+c} \\ 264.722515374 \cdot 1.01980216076^x &=& 1.01980216076^x \cdot 1.01980216076^{c} \\ 264.722515374 &=& 1.01980216076^{c} \qquad | \qquad \ln{()} \\ \ln{(264.722515374)}&=& c\cdot \ln{(1.01980216076)} \\ c &=& \frac{ \ln{(264.722515374)} } { \ln{(1.01980216076)} } \\ c &=& \frac{ 5.57868216559 } { 0.01960864845 } \\ \mathbf{c} & \mathbf{=} & \mathbf{284.501105731 } \\ \\ \hline \\ f(x) &=& b^{x+c} \\ \boxed{~f(x) = 1.01980216076^{x+284.501105731}~} \end{array} $}} \)
\(\small{\text{$ \begin{array}{rcl} f(-5) &=& 1.01980216076^{-5+284.501105731} \\ f(-5) &=& 1.01980216076^{279.501105731} \\ f(-5) &=& 1.01980216076^{279.501105731} \\ f(-5) &=& 240 \qquad \text{okay!} \\\\ f(40) &=& 1.01980216076^{40+284.501105731}\\ f(40) &=& 1.01980216076^{324.501105731}\\ f(40) &=& 1.01980216076^{324.501105731}\\ f(40) &=& 580 \qquad \text{okay!} \\\\ \end{array} $}} \)
+26400 Find a formula for f(x), an exponential function such that f(-5) = 240 and f(40) = 580. f(x)=________
\(\small{ \begin{array}{lrcl} (1) & 580 &=& a\cdot b^{40} \\ (2) & 240 &=& a\cdot b^{-5} \\ \\ \hline \\ (1) : (2) & \frac{580}{240} &=& \frac{ b^{40} }{ b^{-5} } \\ & \frac{580}{240} &=& b^{40}\cdot b^{-5}\\ & \frac{580}{240} &=& b^{45}\\ & \frac{58}{24} &=& b^{45}\\ & \frac{29}{12} &=& b^{45}\\ & b &=& \sqrt[45]{\frac{29}{12}} \\ & b &=& \sqrt[45]{2.416\overline{6}} \\ & b &=& 2.416\overline{6}^{~(0.02\overline{2})}\\ & \mathbf{b} & \mathbf{=} & \mathbf{1.01980216076} \\ \\ \hline \\ & a &=& \frac{580} { b^{40} } \\ & a &=& \frac{580} { 2.19097343941 }\\ & \mathbf{a} & \mathbf{=} & \mathbf{264.722515374} \\ \\ \hline \\ & f(x) &=& a \cdot b^x \\ & f(x) &=& 264.722515374 \cdot 1.01980216076^x \end{array} }\)
\(\small{ \text{A function of the form }\\ f(x) = a\cdot b^x \\ \text{ can be re-written as } \\ f(x) = b^{x ± c} \\ \text{ by the use of logarithms and so is an exponential function.} } \)
\(\small{\text{$ \begin{array}{rcl} 264.722515374 \cdot 1.01980216076^x &=& 1.01980216076^{x+c} \\ 264.722515374 \cdot 1.01980216076^x &=& 1.01980216076^x \cdot 1.01980216076^{c} \\ 264.722515374 &=& 1.01980216076^{c} \qquad | \qquad \ln{()} \\ \ln{(264.722515374)}&=& c\cdot \ln{(1.01980216076)} \\ c &=& \frac{ \ln{(264.722515374)} } { \ln{(1.01980216076)} } \\ c &=& \frac{ 5.57868216559 } { 0.01960864845 } \\ \mathbf{c} & \mathbf{=} & \mathbf{284.501105731 } \\ \\ \hline \\ f(x) &=& b^{x+c} \\ \boxed{~f(x) = 1.01980216076^{x+284.501105731}~} \end{array} $}} \)
\(\small{\text{$ \begin{array}{rcl} f(-5) &=& 1.01980216076^{-5+284.501105731} \\ f(-5) &=& 1.01980216076^{279.501105731} \\ f(-5) &=& 1.01980216076^{279.501105731} \\ f(-5) &=& 240 \qquad \text{okay!} \\\\ f(40) &=& 1.01980216076^{40+284.501105731}\\ f(40) &=& 1.01980216076^{324.501105731}\\ f(40) &=& 1.01980216076^{324.501105731}\\ f(40) &=& 580 \qquad \text{okay!} \\\\ \end{array} $}} \)
