Solving an exponential function

In summary: Yes, very close to it.Yes, very close to it.yes, ##x=4.00 ## is an approximate solution to the problem.yes, ##x=4.00 ## is an approximate solution to the problem.
  • #1
chwala
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Homework Statement


[/B]
Solve the equation
## e^{2x}+2=e^{3x-4}##

Homework Equations

The Attempt at a Solution


I know by using Newton-Raphson method the problem can be solved, i however tried solving it as follows
##e^{3x-4}-e^{2x}-2=0, e^{3x}-e^{2x}.e^{4}-2e^{4}=0, p^3-p^2.e^4-2e^4=0, →p^3-54.6p^2-109.2=0##
where
##p=e^x##,
am i correct, will i get solution this way
 
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  • #2
chwala said:

Homework Statement


[/B]
Solve the equation
## e^{2x}+2=e^{3x-4}##

Homework Equations

The Attempt at a Solution


I know by using Newton-Raphson method the problem can be solved, i however tried solving it as follows
##e^{3x-4}-e^{2x}-2=0, e^{3x}-e^{2x}.e^{4}-2e^{4}=0, p^3-p^2.e^4-2e^4=0##
where
##p=e^x##,
am i correct, will i get solution this way
Yes, you can do it.
 
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  • #3

ehild said:
Yes, you can do it.
i will post solution soon, i don't have calculator now but i know solution will be of the form ## x=ln p##
 
  • #4
the solution is thus ##p=54.64→x=ln 54.64, ⇒x=4.00##
 
  • #5
chwala said:
the solution is thus ##p=54.64→x=ln 54.64, ⇒x=4.00##
Yes, very close to it.
 
  • #6
ehild said:
Yes, very close to it.
yes, ##x=4.00 ## is an approximate solution to the problem.
 
  • #7
chwala said:
yes, ##x=4.00 ## is an approximate solution to the problem.
Might there be other approximate solutions to the problem, i ignored the negative values of ##p=-0.0183+1.41i, -0.0183-1.41i##
 
  • #8
chwala said:

Homework Statement


[/B]
Solve the equation
## e^{2x}+2=e^{3x-4}##

Homework Equations

The Attempt at a Solution


I know by using Newton-Raphson method the problem can be solved, i however tried solving it as follows
##e^{3x-4}-e^{2x}-2=0, e^{3x}-e^{2x}.e^{4}-2e^{4}=0, p^3-p^2.e^4-2e^4=0, →p^3-54.6p^2-109.2=0##
where
##p=e^x##,
am i correct, will i get solution this way

You will get an approximation. To get better accuracy, you need a more accurate value for ##e^4##. In principle you can write down an exact solution, using exact formulas for solutions of a cubic equation and retaining ##e^4## in symbolic form.
 
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  • #9
chwala said:
the solution is thus ##p=54.64→x=ln 54.64, ⇒x=4.00##

You can see that ##x=4## cannot be the solution, but is very close to a solution. For ##x = 4## the left-hand-side is ##2+e^{8\times 4} = 2 + e^8## while the right-hand-side is ##e^{3 \times 4 - 4} = e^8##. Since ##e^8 \doteq 2980.958## the two sides of the equation are close in terms of relative magnitude (that is, % difference), although they still differ by 2 in absolute terms.

You can also see how to get a good approximation quickly: set ##x = 4 + y## with ##y## small. The equation becomes ##2 + e^8 e^{2y} = e^8 e^{3y}##, or ##e^{3y} - e^{2y} = 2 e^{-8} \doteq##0.6709252558e-3. Using the series expansions ##e^{3y} = 1 + 3y + \cdots## and ##e^{2y} = 1 + 2y + \cdots##, we have ##3y - 2y + \cdots =##0.6709252558e-3, so ##y \doteq##0.6709252558e-3 and ##x \doteq 4.000670925.## This is still an approximation, but it is a pretty good one: with this new ##x## the left-hand side is 2986.960670, while the right-hand-side is 2986.964042.
 
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  • #10
Ray Vickson said:
You can see that ##x=4## cannot be the solution, but is very close to a solution. For ##x = 4## the left-hand-side is ##2+e^{8\times 4} = 2 + e^8## while the right-hand-side is ##e^{3 \times 4 - 4} = e^8##. Since ##e^8 \doteq 2980.958## the two sides of the equation are close in terms of relative magnitude (that is, % difference), although they still differ by 2 in absolute terms.

You can also see how to get a good approximation quickly: set ##x = 4 + y## with ##y## small. The equation becomes ##2 + e^8 e^{2y} = e^8 e^{3y}##, or ##e^{3y} - e^{2y} = 2 e^{-8} \doteq##0.6709252558e-3. Using the series expansions ##e^{3y} = 1 + 3y + \cdots## and ##e^{2y} = 1 + 2y + \cdots##, we have ##3y - 2y + \cdots =##0.6709252558e-3, so ##y \doteq##0.6709252558e-3 and ##x \doteq 4.000670925.## This is still an approximation, but it is a pretty good one: with this new ##x## the left-hand side is 2986.960670, while the right-hand-side is 2986.964042.
Thanks Ray, yeah right I know the solution may be approximated by various numerical methods techniques, I appreciate bro
 
  • #11
chwala said:
Might there be other approximate solutions to the problem? I ignored the negative values of ##p=-0.0183+1.41i, -0.0183-1.41i##.
Those aren't negative roots; they're complex roots.

You had a cubic polynomial, so it has only three roots. The one real root gave you the solution you wanted. The two complex roots won't yield real solutions, so there are no more solutions in the (assumed) domain of the problem.
 
  • #12
Agreed
 

FAQ: Solving an exponential function

1. What is an exponential function?

An exponential function is a mathematical function in the form of f(x) = ab^x, where a and b are constants and x is the variable. It is characterized by a constant ratio between the input (x) and output (f(x)) values, meaning that as x increases or decreases, the output changes at a constant rate.

2. How do you solve an exponential function?

To solve an exponential function, you can use logarithms or graphing techniques. If you are using logarithms, you can use the property log(ab) = log(a) + log(b) to simplify the equation and isolate the variable. If you are graphing, you can plot points and find the pattern to determine the value of the variable.

3. What is the difference between an exponential and a linear function?

An exponential function has a constant ratio between the input and output values, while a linear function has a constant difference between the input and output values. Additionally, exponential functions have a curved graph, while linear functions have a straight line graph.

4. What are some real-life applications of exponential functions?

Exponential functions are commonly used to model population growth, compound interest, and radioactive decay. They can also be used to describe the growth or decline of bacteria, viruses, and diseases.

5. Can you have a negative base in an exponential function?

Yes, you can have a negative base in an exponential function. However, the base must be an odd number for the function to be a one-to-one function. This means that each input (x) value corresponds to a unique output (f(x)) value, and vice versa.

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