Can someone show me how to do this please. I am stuck.
find a number x such that
e^2x – 12e^x =13
Substitute y=e^x to get a quadratic equation:
y²-12x+13=0
Solve for y.
Substitute the values of y=e^x into the original equation to make sure that all the roots are feasible/defined.
See also previous example:
http://www.jiskha.com/display.cgi?id=1291874924
let e^x = t
then
e^2x – 12e^x =13 becomes
t^2 - 12t - 13 = 0
(t-13)(t+1) = 0
t = 13 or t = -1
then e^x = 13 or e^x = -1
x = ln 13 or x = ln(-1)
but in ln(a) a > 0
so x = ln 13
I slipped in the sign of the transformed equation. Instead of
y²-12x+13=0
it should read
y²-12x-13=0
Go with Reiny's solution.
Thank u both
Sure, I can help you with that. The equation you provided is a quadratic equation in terms of e^x. To solve for x, we can use a substitution. Let's substitute e^x with u. So our new equation becomes:
u^2 - 12u = 13
Now we have a regular quadratic equation. To solve for u, we need to bring all terms to one side to form a quadratic equation, set it equal to zero, and then solve using factoring or the quadratic formula.
u^2 - 12u - 13 = 0
Next, we can factor or use the quadratic formula. Let's try factoring:
(u - 13)(u + 1) = 0
To find the values of u, we set each factor equal to zero:
u - 13 = 0 or u + 1 = 0
Solving for u, we get:
u = 13 or u = -1
Since we set u = e^x, we can solve for x by taking the natural logarithm (ln) of both sides:
ln(u) = ln(13) or ln(u) = ln(-1)
Using logarithmic properties, the second equation has no solution because the natural logarithm of a negative number is not defined. However, the first equation has a solution:
ln(13) = x
So, the value of x is ln(13), which is approximately 2.564949.