SOLUTION: I have to solve this problem: 2+6e(4x)=19 (the 4x is the exponent) I have come up with these possible answers: .2604, .5074, .4907, or 2.8333

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I have to solve this problem: 2+6e(4x)=19 (the 4x is the exponent) I have come up with these possible answers: .2604, .5074, .4907, or 2.8333      Log On


   

Question 81497This question is from textbook Glencoe Mathmatics Algebra 2
: I have to solve this problem:
2+6e(4x)=19 (the 4x is the exponent)
I have come up with these possible answers:
.2604, .5074, .4907, or 2.8333 This question is from textbook Glencoe Mathmatics Algebra 2

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
This is a classic problem of "undo-ing" the operations of math in order to solve for x.

2+6e(4x)=19

First, you need to "undo" the 2 that is added to the term that contains the x. Of course, you do this by subtracting 2 from each side of the equation:

6e%5E%284x%29+=+17

Next, you must "undo" the 6 that is multiplied times the term containing the x:
e%5E%284x%29+=+17%2F6

Next, you must "undo" the "e" raised to the power. The inverse of "e" raised to the power is the "ln" function, so you must take the "ln" of each side:
ln%28e%5E%284x%29%29+=+ln%2817%2F6%29
4x=ln%2817%2F6%29

Finally, undo multiplication of the x times 4 by either dividing both sides by 4 or by multiplying both sides times 1%2F4.

Final answer x=+%28ln%2817%2F6%29%29%2F4+
x=+.2604

R^2 at SCC