SOLUTION: if a graph is created showing ln z against t for this : z= 2e^(4t) - what is the gradient, and how do I work it out - I am struggling with understanding a text book. Please note t

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: if a graph is created showing ln z against t for this : z= 2e^(4t) - what is the gradient, and how do I work it out - I am struggling with understanding a text book. Please note t      Log On


   

Question 894150: if a graph is created showing ln z against t for this : z= 2e^(4t) - what is the gradient, and how do I work it out - I am struggling with understanding a text book. Please note this is not the actual problem I need to solve - it is similar - I need to understand how to do this type of question.
I have tried to follow the text and thought the gradient was equal to the power when I multiply by base e and reverse the order etc then use the line equation y=mx+c. Please can you help.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The gradient is the derivative of the function.
In this case,
z=2e%5E%284t%29
dz%2Fdt=2e%5E%284t%29%2A4
dz%2Fdt=8e%5E4t
The value of the tangent line to the function at any point is the value of the derivative at that point.
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.
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As an example, when t=0,z=2,dz%2Fdt=8
So the equation of the tangent line is,
z-2=8%28t-0%29
z=8t%2B2
On the graph, x=t and y=z
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.
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graph%28300%2C300%2C-2%2C2%2C-1%2C5%2C2e%5E%284x%29%2C8x%2B2%29