A new kind of rocket takes off along an exponential trajectory, with height, in miles, represented by 3x, where x is the time, in seconds. Find the time when the height of the rocket is 8 miles.

What is the exact solution written as a logarithm?

What is an approximate solution rounded to the nearest thousandth?

Respuesta :

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.

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