What value of x do the graphs of f and g have parallel tangent lines?

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SUMMARY

The problem involves finding the value of x where the functions f(x) = 3e2x and g(x) = 6x3 have parallel tangent lines. The correct value of x is -0.391, determined by equating the derivatives f'(x) = 6e2x and g'(x) = 18x2. The solution approach includes setting the derivatives equal and suggests using substitution of given answer choices to find the correct x value, rather than solving the equation algebraically.

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Homework Statement



Let f be the function given by f(x) = 3e^2x and let g be the function given by g(x) = 6x^3. At what value of x do the graphs of f and g have parallel tangent lines?

a. -0.701
b. -0.567
c. -0.391
d. -0.302
e. -0.258

Correct answer is c. -0.391


Homework Equations





The Attempt at a Solution



Well, tangent line means doing the derivative. Thus, I did the derivative of f and g.

f'(x) = 6e^2x
g'(x) = 18x^2​

Parallel means the slopes or the derivatives are the same. Thus, I set them equal to each other in order to solve for x.

f'(x) = g'(x)
6e^2x = 18x^2
e^2x = 3x^2​

Here I was stuck. I could do ln(e^2x) to cancel out the e, but I would end up with something ugly on the right side.

2x = ln(3x^2)​

and I wouldn't have x on one side. Am I approaching this problem incorrectly, or is my algebra wrong?
 
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Your method is correct. But to fully solve it you either need to graph it or use an iterative method. Instead, since you are given answers and you know that f'(x)=g'(x), why not just sub in the values for x and see which one produces the result f'(x) being the same as g'(x).
 

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