I have a quiz today on logarithms and need help!?

Question

I have some sample problems to study with, but I can't figure this one out.

Solve this equation:
e^x-12e^-x -1=0

Please show me the steps.
Thanks.

I made an error in the problem; it should be:

(e^x) - (12e^-x) - (1) = 0

I received a response--which I entirely appreciate--but I still don't understand. If someone could explain the steps, too, that'd help a whole lot!

Answer 1

Let y = e^x and recall that e^-x == 1/(e^x).
Now our equation becomes
y - 12(1/y) - 1 = 0
That is a quadratic equation "in disguise". To solve it, multiply by y throughout:
y^2 - 12 - y = 0
y^2 + (-1) y + (-12) = 0
Solve this for y (you should be able to factorise this one by inspection).
Recall that e^x can't possibly be negative, since e itself is positive. This means one of the solutions to the quadratic can be rejected.
Take the natural logarithm of whichever of the two solutions is positive.
That's your x.

Answer 2

e^x - 12/e^x - 1 = 0 clear the denominator by multiplying every term by e^x
e^2x - 12 - 1e^x = 0
e^2x - e^x - 12 = 0
Factoring
(e^x -4)(e^x + 3) = 0
e^x = 4 or e^x = -3 impossible to = -3 so stick with the 4 answer

e^x = 4
take the ln of both sides
ln e^x = ln 4 The left simplifies to just x by properties of ln
x = ln 4

Answer 3

this is amar
(e^x) - (12e^-x) - (1) = 0
(e^x)-1=(12e^x)
(12e^-x)+(e^x)=1
(13e^x)=1
(e^x)=1/13
(e^x) - (12e^-x) - (1) = 0
(1/13)-(12*1/13)-(1)=0

(1/13)-(12/13)=-1
(0.0769230769)-(0.923076923)= -1
0.0769230769+0.923076923= -1
1= -1
1-1=0
0=0

Answer 4

e^x - 12e^-x - 1 = 0


Let e^x be y,

y - 12/y - 1 = 0
y^2 - 12 - y = 0 (Multiply equation by y)
y^2 - y - 12 = 0
(y - 4)(y + 3) = 0

Hence,
y - 4 = 0 --------------or ------------- y + 3 = 0
y = 4 -------------------or-------------- y = -3 (rejected)

Hence,
e^x = 4
ln e^x = ln 4 (ln both sides)
x ( ln e ) = ln 4
x= ln 4 / ln e
=1.39 ( 3.sig.fig)

Answer 5

(e^x)-(12e^-x)-(1)=0
e^x-(12/e^x)-1=0(since e^-x =1/e^x)
taking l.c.m
e^2x-12-e^x=0
e^2x-e^x=12
applying log on both sides
2x-x=log12
x=log12

Answer 6

(e^x) - (12e^-x) - (1) = 0
(e^x)-1=(12e^x)
(12e^-x)+(e^x)=1
(13e^x)=1
(e^x)=1/13