How Do You Solve \( e^{x-1} = 5 - y^2 + y \) for x?

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To solve the equation \( e^{x-1} = 5 - y^2 + y \) for x, the first step involves taking the natural logarithm, resulting in \( x - 1 = \ln(5 - y^2 + y) \). This leads to the solution \( x = \ln(5 - y^2 + y) + 1 \). However, it is crucial to note that the expression \( 5 - y^2 + y \) must be greater than zero, imposing a constraint on the values of y. A minor correction was made regarding the expression for x, confirming it should be \( x = \ln(..) + 1 \). The discussion emphasizes the importance of considering the inequality for valid y values.
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Homework Statement


Solve e x-1 = 5-y2+y for x

Homework Equations


/

The Attempt at a Solution


ln(ex-1)=ln(5-y2+y)
x-1=ln(5-y2+y)
x=ln(5-y2+y)+1

can I go any further ?
 
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Nope, just mind one thing, you have to include a constraint that 5-y2+y>0 which holds only if y satisfies an inequality regarding its values.

Btw i just noticed one small "typo", it should be x=ln(..)+1 not -1...
 
Delta² said:
Nope, just mind one thing, you have to include a constraint that 5-y2+y>0 which holds only if y satisfies an inequality regarding its values.

Btw i just noticed one small "typo", it should be x=ln(..)+1 not -1...
thank you very much for the help man :)
oops fixed that typo
 

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