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IB Math Methods- Problems with Problems!

I have a few problems:
Solve
1.
32^2x = e
e^2x +1

I cross multiplied and got 3e^2x = e(e^2x + 1) Then:
3e^2x = e^2x+e And now I don't know what to do from here....

2. Solve

e^2x - 5e+x + 6 = 0

This one can't be factored, so does the quadratic equation work here? If so, how is it done with "e"????
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,845
17
First, I'd double check and make sure you wrote both of these equations correctly. I think you've made at least two typos. Also, you have to group your exponents with parentheses. i.e.:

e^2x = e2x = e2 * x
but
e^(2x) = e2x

People will still understand what you mean, but you should always strive to write things correctly instead of relying on people to read your mind. Generally, you should avoid ever writing e^2x, even if you are using it correctly, because it can generate confusion.



Some things to consider for the first problem:

Did you distribute right?

Can you solve the equation for e2x?


For the second problem:

Once you get the right statement of the problem, consider that e2x = (ex)2
 
Last edited:

HallsofIvy

Science Advisor
Homework Helper
41,712
876
Like Hurkyl, I would say that e^2x is almost certainly intended to be e^(2x), but I would not be nearly so sure about e^2x+1 !

Is that intended to be e^(2x)+ 1 or e^(2x+1)? I would say either was likely and they are very different.

IF it were 3e^(2x)/(e^(2x+1)) = e, the left side reduces to
3e^(2x-(2x+1))= 3 e^(-1) which is not e so I guess that was not intended!

3e^(2x)/(e^(2x)+1)= e gives, after you multiply both sides by that denominator, 3e^(2x)= e(e^(2x)+1)= e*e^(2x)+ e. Now, like you learned to do long ago, subtract that e*e^(2x) from both sides to get x on only one side of the equation.

You should now have 3e^(2x)- e e^(2x)= (3-e)e^(2x)= e.

If you have an equation of the form Ax= B, you would divide both sides by A wouldn't you? Okay, divide both sides of the equation by
3-e to get e^(2x)= e/(3-e).

Now the "new" part. How do you get rid of that "e^ " part?
Well, natural logarithm, ln(x) is DEFINED as the inverse (opposite) of e^x. What happens if you take ln of both sides of the equation?
 

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