Exponential Function

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NobodysHero

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Using the formula a^m/n = (a^1/n)^m = nth root of (a^m) = (nth root of (a))^m. Solve e^(3-x) = (e^3)^-x. I have no idea how to solve this problem can someone teach me how to solve equations like this?
 
NobodysHero said:
Using the formula a^m/n = (a^1/n)^m = nth root of (a^m) = (nth root of (a))^m. Solve e^(3-x) = (e^3)^-x. I have no idea how to solve this problem can someone teach me how to solve equations like this?
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Do you recognize that e^(3- x)= e^3(e^(-x))? And that (e^3)^(-x)= (e^(-x))^3? What I would do is let y= e^(-x) so that equation becomes (e^3)y= y^3. Since y= e^(-x), y cannot be equal to 0 so you can divide both sides by y to get y^2= e^3. Now take the square root of both sides and you that formula to write it as e^ a fraction.
 
I was looking at a few of the examples in the book and it doesn't include y at all. If I'm understanding the examples correctly, The equation: e^(3-x) = (e^3)^(-x) have the same base. So the next step would be to solve 3-x = 3-x. I then subtract 3 from both sides and add x to both sides which results in 0=0. That doesn't seem right but looking at the examples those are the steps that were used.
 
NobodysHero said:
I was looking at a few of the examples in the book and it doesn't include y at all. If I'm understanding the examples correctly, The equation: e^(3-x) = (e^3)^(-x) have the same base. So the next step would be to solve 3-x = 3-x. I then subtract 3 from both sides and add x to both sides which results in 0=0. That doesn't seem right but looking at the examples those are the steps that were used.
Click to expand...

\(\displaystyle e^{3-x} \ = \ \left [e^3\right]^{-x}\)

3-x = - 3*x

3 = -3*x + x

x = -3/2

There are many ways to solve the problem. Since you did not show any work - we did not know what you knew.

Post your work along with the problem - you'll get faster help.
 
NobodysHero said:
Using the formula a^m/n = (a^1/n)^m = nth root of (a^m) = (nth root of (a))^m. Solve e^(3-x) = (e^3)^-x. I have no idea how to solve this problem can someone teach me how to solve equations like this?
Click to expand...
I think this problem as presented is about as confusing as is imaginable. I do not see where the formula given leads to an intuitively appealing solution. Consequently, I am going to do what I rarely do, namely take you through the problem in a more intuitive way.

\(\displaystyle Given\ a > 0\ and\ a\ne 1, then\ a^u = a^v \iff u = v.\)

Does that make sense to you?

\(\displaystyle e^{(3 - x)} = \left(e^3\right)^{-x} \implies e^{(3 - x)} = e^{-3x}.\) Just applied one of the laws of exponents.

\(\displaystyle So\ 3 - x = - 3x.\) An application of the general rule shown above.

Still with me?

\(\displaystyle So\ 3 = x - 3x = -2x \implies x = -\dfrac{3}{2}.\)

Now check your work

\(\displaystyle 3 - x = = 3 - \left(-\dfrac{3}{2}\right) = 3 + \dfrac{3}{2} = \dfrac{6 + 3}{2} = \dfrac{9}{2}.\)

\(\displaystyle 3 * (- x) = (-3) * x = (-3) * \left(- \dfrac{3}{2}\right) = 3 * \dfrac{3}{2} = \dfrac{9}{2}.\)

And quite obviously \(\displaystyle e^{(9 \div 2)} = e^{(9 \div 2)}.\) It checks.

Halls of Ivy has shown how to use the formula, but that is not intuitive to me at all.
 
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JeffM that's exactly what I needed. Thank you for that. My apologizes to HallsofIvy I should've added the steps there to clarify which way it needed to be solved. Thank you guys for all your help.
 
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