SOLUTION: please help me solve this using substitution y=3^X, solve the equation 3^x=2+3^-x.

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Question 1075860: please help me solve this
using substitution y=3^X, solve the equation 3^x=2+3^-x.
Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(52790)   (Show Source): You can put this solution on YOUR website!
.
O, great !!! This is entirely different and fully correct request. 

After substituting  y = , your equation becomes

y = 

Multiply by "y" both sides. You will get

 = ,  or

 = 0.


Apply the quadratic formula to solve the quadratic equation. You will get

 =  =  = .


The negative root does not make sense, since y =  must be positive. 
It is EXTRANEOUS root for our problem.

The only positive root y =  survives.


So, we have  = , which implies

x = .

Solved.



Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
solve the equation 3^x=2+3^-x.
----------
3^x = 2 + 3^-x
3^x - 2 - 3^-x = 0
Mutliply thru by 3^x
3^(2x) - 2*3^x - 1 = 0
Sub y for 3^x
y^2 - 2y - 1 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=8 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 2.41421356237309, -0.414213562373095. Here's your graph:

==============
Ignore the negative solution
3^x = 1 + sqrt(2)
==========
x*log(3) = log(1 + sqrt(2))
x = log(1 + sqrt(2))/log(3)
x =~ 0.80226
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