SOLUTION: Using suitable substitutions solve for x, 2e^x=7√e^x -3

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Question 1147557: Using suitable substitutions solve for x, 2e^x=7√e^x -3
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52795) About Me  (Show Source):
You can put this solution on YOUR website!
.

            Look into your post, please, as it came to the forum.

            From the post, it is UNCLEAR how far the square root symbol is spreading to the right. 


If your equation is   


    2%2Ae%5Ex = 7%2Asqrt%28e%5Ex%29 - 3      (1)


then the solution is as follows.



Introduce new variable  u = sqrt%28e%5Ex%29.


Then the equation (1) takes the form


    2u%5E2 = 7*u - 3

    2u%5E2+-+7u+%2B+3 = 0

    (2u-1)*(u-3) = 0


Case 1.  2u - 1 = 0

                 u = 1%2F2

                sqrt%28e%5Ex%29 = 1%2F2

                e%5Ex = 1%2F4

                x = ln%281%2F4%29 = -2*ln(2).  

                It is the answer for Case 1.



Case 2.  u - 3 = 0

                 u = 3

                sqrt%28e%5Ex%29 = 3

                e%5Ex = 9

                x = ln%289%29%29 = 2*ln(3).  

                It is the answer for Case 2.


ANSWER.  The given equation has two solutions,  -2*ln(2)  and  2*ln(3).


Solved.

--------------

The solution by @greenestamps has editorial deficiencies,

so I came to fix them.


Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Let u+=+e%5E%28sqrt%28x%29%29

Then the equation is

2u%5E2+=+7u-3
2u%5E2-7u%2B3+=+0
%282u-1%29%28u-3%29+=+0

u+=+1%2F2 or u+=+3

(1) e%5E%28sqrt%28x%29%29+=+1%2F2
e%5Ex+=+1%2F4
x+=+ln%281%2F4%29

(2) e%5E%28sqrt%28x%29%29+=+3
e%5Ex+=+9
x+=+ln%289%29