Solve the equation e^x+9e^-x/2=3 What is the solution in terms of natural logarithms? x= What is the decimal approximation for the solution? x=
It depends on what you mean. If you mean \(\dfrac{e^{x} + 9e^{-x}}{2} = 3\), then you should write that. You definitely have not.
Yes, that is what I meant sorry...
Do you now why that is not what you have written?
Refer to the attachment generated by Wolfram Alpha
Since I was a little crabby, I'll give you one for free... \(\dfrac{e^{x} + 9e^{-x}}{2} = 3\) 2 is never zero. Multiply by it. \(e^{x} + 9e^{-x} = 6\) \(e^{x}\) is never zero. Multiply by it \(e^{2x} + 9 = 6e^{x}\) Rearrange a little. \(e^{2x} - 6e^{x} + 9 = 0\) Recognize this as a quadratic-looking structure and simply factor it. \((e^{x} - 3)^{2} = 0\) Can you do the rest?
So now simplify it?
Is that the solution? How do I make it into a decimal?
You cannot simply that. We're solving aren't we? Do we have a value for 'x'? That would be a solution.
If you were solving a quadratic equation and encountered this, \((x-3)^{2} = 0\), what would you do?
I am guessing some how put it into decimal form
You're saying that if you factor a quadratic equation, and get \((x-3)^{2} = 0\), you cannot solve it? x = What?
Since I am having to put it into decimal terms and find the solution
3
How did you get that?
Solved the equation for x
Is it wrong
Given \((x-3)^{2} = 0\), we conclude that \(x - 3 = 0\) or \(x = 3\). Is that what you mean by "solved the equation"?
yes
How would I put it into decimal form though?
Perfect. Now, do EXACTLY the same thing with \((e^{x} - 3)^{2} = 0\)
decimal approximation**
We're not there, yet. Let's find the EXACT solution.
Remember that quadratic equation. Apply that EXACT technique to this equation.
x=ln(3)
or 1.9086
It helps a conversation if you go one step at a time. Very good. The REAL trick here was to recognize the Quadratic Form and treat it like that.
Ok so which one is the exact solution and do I put 1.9086 for the decimal
What is not exact about ln(3)?
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