We have been given that a new kind of rocket takes off along an exponential trajectory, with height, in miles, represented by [tex]3^x[/tex], where x is the time, in seconds. We are asked to find the time when the height of the rocket id 8 miles.
To find the time, we will equate height function with 8 and solve for x as:
[tex]3^x=8[/tex]
To solve for x, we will take natural log on both sides as:
[tex]\ln(3^x)=\ln (8)[/tex]
We can rewrite 8 as [tex]2^3[/tex].
[tex]\ln(3^x)=\ln (2^3)[/tex]
Using natural log property [tex]\ln(a^b)=b\cdot \ln(a)[/tex], we will get:
[tex]x\cdot \ln(3)=3\cdot\ln (2)[/tex]
[tex]\frac{x\cdot \ln(3)}{ \ln(3)}=\frac{3\cdot\ln (2)}{ \ln(3)}[/tex]
[tex]x=\frac{3\cdot\ln (2)}{ \ln(3)}[/tex]
Therefore, exact solution will be [tex]\frac{3\cdot\ln (2)}{ \ln(3)}[/tex].
[tex]x=\frac{2.0794415416798359}{1.0986122886681097}[/tex]
[tex]x=1.892789260714[/tex]
Upon rounding to nearest thousandth, we will get:
[tex]x\approx 1.89[/tex]
Therefore, the height of the rocket will be 8 miles after approximately 1.89 seconds.