SOLUTION: Having trouble with this question: Solve for x. e is a number, not a variable 2e^(x-2) = e^x + 7 So far, here is my work. I keep going in circles: e^(x-2) = (e^x + 7)/2 ln

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Having trouble with this question: Solve for x. e is a number, not a variable 2e^(x-2) = e^x + 7 So far, here is my work. I keep going in circles: e^(x-2) = (e^x + 7)/2 ln      Log On


   

Question 185080This question is from textbook Prentice Hall Mathematics Algebra 2
: Having trouble with this question:
Solve for x. e is a number, not a variable
2e^(x-2) = e^x + 7
So far, here is my work. I keep going in circles:
e^(x-2) = (e^x + 7)/2
ln(e^(x-2)) = ln((e^x + 7)/2)
x-2 = ln(e^x + 7) - ln(2)
I get lost here. This question is from textbook Prentice Hall Mathematics Algebra 2

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
2e%5E%28x-2%29+=+e%5Ex+%2B+7+ Start with the given equation.


2e%5E%28x%29%2Fe%5E2+=+e%5Ex+%2B+7 Rewrite e%5E%28x-2%29 as e%5E%28x%29%2Fe%5E2 using the identity x%5E%28y-z%29=x%5Ey%2Fx%5Ez where x%3C%3E0


2e%5E%28x%29+=+e%5Ex%2Ae%5E2+%2B+7%2Ae%5E2 Multiply EVERY term by the LCD e%5E2 to get rid of the fraction.


2e%5E%28x%29+-+e%5Ex%2Ae%5E2+=+7%2Ae%5E2 Subtract e%5Ex%2Ae%5E2 from both sides.


e%5E%28x%29%282+-+e%5E2%29+=+7%2Ae%5E2 Factor out the GCF e%5Ex


e%5E%28x%29+=+7%2Ae%5E2%2F%282+-+e%5E2%29 Divide both sides by 2+-+e%5E2.


e%5E%28x%29+=+-9.597 Evaluate the right side with a calculator.


x+=+ln%28-9.597%29 Take the natural log of both sides (to eliminate the base "e")


Since you CANNOT take the natural log of a negative number (well at least for now...), this means that there is no solution.



So there are no solutions. I would double check the original problem.


As visual confirmation, graph the two expressions 2e%5E%28x-2%29 and e%5Ex+%2B+7 and you'll find that they do NOT cross.



Note: taking the log of a negative number will give you a complex number, but that's extra information...