How do you solve for x in the equation: e^(2x)-(7x)+10=0?

[xhopsx]

I do not usually need help, but I have been stuck on this problem for over a week.
I am trying to solve the problem without using a calculator. I know I can show the answer graphically, but that does not really give me a valid answer for my instructor. Ideas that I have been playing around with/tried:
Substituting a value for a value, such as "r"- I do not know what value to substitute
Raising everything by e- I was desperate when trying this and got no where.
having 2x=ln(7x-10)- I get stuck here. I do not know where to go on from this point. I have also tried doing a domain for ln(7x-10) but am not sure how to use it.
I have tried a few other ideas, but I am not sure what to do with them.

 

[lookagain]

I do not usually need help, but I have been stuck on this problem for over a week.
I am trying to solve the problem without using a calculator. I know I can show the answer graphically,
but that does not really give me a valid answer for my instructor. Ideas that I have been playing around with/tried:
Substituting a value for a value, such as "r"- I do not know what value to substitute
Raising everything by e- I was desperate when trying this and got no where.
having 2x=ln(7x-10)- I get stuck here. I do not know where to go on from this point.
I have also tried doing a domain for ln(7x-10) but am not sure how to use it.
I have tried a few other ideas, but I am not sure what to do with them.

xhopsx, is there any chance that the equation is actually e^(2x) - 7e^x + 10 = 0?

In Latex that is \(\displaystyle e^{2x} - 7e^x + 10 = 0\)


If that were so, that equation could be manipulated to (e^x - 2)(e^x - 5) = 0.

 

[DrPhil]

I hope lookagain is right about the problem! But if not, the only way to solve is numerically.

I do not usually need help, but I have been stuck on this problem for over a week.
I am trying to solve the problem without using a calculator. I know I can show the answer graphically, but that does not really give me a valid answer for my instructor. Ideas that I have been playing around with/tried:
Substituting a value for a value, such as "r"- I do not know what value to substitute
Raising everything by e- I was desperate when trying this and got no where.
having 2x=ln(7x-10)- I get stuck here. I do not know where to go on from this point. I have also tried doing a domain for ln(7x-10) but am not sure how to use it.
I have tried a few other ideas, but I am not sure what to do with them.

I would go with the logarithm.

Let \(\displaystyle y(x) = 2x - \ln{(7x - 10)} \)

..\(\displaystyle \dfrac{dy}{dx} = 2 - \dfrac{7}{7x - 10} = 2 - \dfrac{1}{x - 10/7}\)

The original equation is solved if y(x) = 0. If you can find a value of x for which y(x)>0 and another value of x for which y(x)<0, then the solution is between, and you can use bifurcation to get as close to the value as you need to be. Setting the derivative to 0 shows that y(x) has either a min or a max at x=1.92857....
Plugging that x into the equation gives
........y(1.9286) = 2.604 > 0
Then looking at values close on either side, I find that that value of y is its minimum.
There is NO value of x that makes y 0, hence no solution of the equation as you wrote it.

So lets hope that lookagain guessed right!

 

[xhopsx]

So, finally my instructor has emailed me back. It is actually 7e^x
This makes much more sense, and is much more easily solved. Thanks for the help!