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

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) (Show Source): You can put this solution on YOUR website!
Start with the given equation.

Rewrite as using the identity where

Multiply EVERY term by the LCD to get rid of the fraction.

Subtract from both sides.

Factor out the GCF

Divide both sides by .

Evaluate the right side with a calculator.

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 and 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...

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