Solving for h(t) Greater Than 3: Is There a Better Way?

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To determine the intervals where h(t) = ln(1+t) + (3/4)*cos(t/2) is greater than 3, the inequality ln(1+t) + (3/4)*cos(t/2) > 3 must be solved. The equation ln(1+t) + (3/4)*cos(t/2) = 3 cannot be solved algebraically due to the presence of transcendental functions. The graphing calculator's "intersect" functionality is a valid method for finding solutions. Further research into transcendental functions is suggested for deeper understanding. This approach highlights the complexity of solving such inequalities analytically.
uchihajeff
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Hi

This is a problem on a practice test.



Function h(t)=ln(1+t) + (3/4)*cos(t/2)

What are the intervals in which h(t) is greater than 3?



So far, the only way I've been able to figure it out is using the "intersect" functionality of my graphing calculator.

Is there a way to isolate the t? Is there a better way to solve this problem, possibly without the calculator?

Thanks
~uchihajeff
:-p
 
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Essentially, then, you want to solve the inequality
ln(1+t) + (3/4)*cos(t/2)> 3 which is equivalent to solving the equation
ln(1+t) + (3/4)*cos(t/2)= 3.

Since that involves two different transcendental functions, there is no algebraic way of solving that equation.
 
Thanks for clearing that up. :smile: Now I'll go research into what transcendental functions are.
 
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