SOLUTION: The substitution u=2^x can be used to convert the equation 4^x−2^(x+1)−15=0 to the form u^2+au+b=0 Ho2 do I find the value of a and b?

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: The substitution u=2^x can be used to convert the equation 4^x−2^(x+1)−15=0 to the form u^2+au+b=0 Ho2 do I find the value of a and b?      Log On


   

Question 1009860: The substitution u=2^x can be used to convert the equation 4^x−2^(x+1)−15=0
to the form u^2+au+b=0
Ho2 do I find the value of a and b?
Found 2 solutions by josmiceli, Boreal:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+4%5Ex+-+2%5E%28+x%2B1+%29+-+15+=+0+
+u+=+2%5Ex+
+u%5E2+=+%28+2%5Ex+%29%5E2+
+u%5E2+=+%28+2%5E2%29%5Ex+
+u%5E2+=+4%5Ex+
------------------
+2%5E%28+x%2B1+%29+=+2%5Ex%2A2%5E1+
+2%5E%28+x%2B1+%29+=+2%2A2%5Ex+
+2%5E%28+x%2B1+%29+=+2u+
-------------------
So, the equation is:
+u%5E2+-2u+-+15+
If this is the form to get it into:
+u%5E2+%2B+au+%2B+b+=+0+
+a+=+-2+
+b+=+-15+

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
u^2+au+b=0
u=2^x
4^x=2^2x=2^x*2^x=u^2
2^(x+1)=2^x*2=2u
u^2-2u-15=0
(u-5)((u+3)=0
u=5, -3, the -3 root is extraneous.
2^x=5
log both sides
xlog 2=log5
x= log 5/log2=2.3219
4^2.3219=25.00
2^3.3219=10
a=-2, b=-15