How do I solve exponentials with base e?

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Homework Help Overview

The discussion revolves around solving equations involving exponentials with base e, specifically focusing on the equation x^3 + e^(2x) + 8 = 0. Participants are exploring methods to find solutions for x in such equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of logarithms to solve equations involving e, questioning how to apply logarithmic properties effectively. Some suggest that exact solutions may not be possible, leading to graphical or iterative methods instead.

Discussion Status

There is an ongoing exploration of different methods to approach the problem, including the application of logarithmic identities. Some participants have provided guidance on using natural logarithms, while others express uncertainty about finding exact solutions and consider alternative methods.

Contextual Notes

Participants note that the original problem may not yield an exact solution and that iterative or graphical methods might be necessary. There is also a mention of the need for foundational knowledge in logarithms to proceed with the discussion.

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I don't know how to solve x for things like e^x=20.
The problem I have to solve is x^3+e^(2x)+8=0
Can anyone help, please?
 
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have you learned logarithms?

what is the natural log of e^x?
 
LinkMage said:
I don't know how to solve x for things like e^x=20.
The problem I have to solve is x^3+e^(2x)+8=0
Can anyone help, please?

You won't be able to find an exact solution to your problem so you will have to resort to solving it graphically or by iteration.
 
The exponential function is the inverse function of the natural logarithm.
therefore [tex]ln(e^{f(x)}) = f(x)[/tex]. In your example, like any equation, what you do on one side of the equation must be done on the other side. You can take the natural log of both sides [tex]ln(e^{x})=ln(20)[/tex]
Then using the property I gave above you should be able to solve for x

Once you have practised using the idea given above, you can then tackle your second question. You may want to bring your x-terms on one side of equation and any other terms, to the other side of equation. As with the first query, take the natural log of both sides. Though you eliminate your exponential function, you may still have some natural log terms left. That's okay, by substitution of values for x you should arrive at correct value f(x) to your question.

Edit: By plotting the values of x you substitute and resulting values f(x), as Tide suggests; by observing the trend in your graph, you may find (or at least narrow down) your solutions more quickly.
 
Last edited:
You mean I just replace values for x until I get that the equation equals 0. That's what I did first and got aproximately -2.001. I was just wondering if I could solve the equation instead of replacing numbers.
 
Anytime you have a logarithm, there are different rules that you need to know. For the first example you gave the one you need to know is this.

[tex]\log_{a}b^c=c\log_{a}b[/tex] it also follows that

[tex]\ln a^b=b\ln a[/tex]

So when you are given something like [tex]e^x=20[/tex], how can you apply this rule to solve that?

For your second problem, as others have said, you cannot solve for x explicity so another method must be used.
 
zwtipp05 said:
have you learned logarithms?

what is the natural log of e^x?
it is equal to x,

well, I just wanted to remind theOP
 

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