How do you solve log base 2 of 0.5 power 3 ?
1 answer:
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Hello,
y=2^(-x)
y=2^(2x)+3
==>2^(2x)+3=1/2^x
==>2^(3x)+3*2^x-1=0 (1)
Let's assume u=2^x
(1)==>u^3+3*u-1=0
which as 3 roots
u=0.322185354626 or
u = -0.161092677313 + i1.754380959784 or
u = -0.161092677313 - i1.754380959784.
Let's take the real solution
0.322185354626=2^x
==>x=ln(0.322185354626) / ln(2)
x=-1,6340371790199...
an other way is
f(x)=2^(3x)+3*2^x-1
f(-2)=1/64+3/4-1=-15/64 <0
f(-1)=1/8+1-1=1/8>0
==> there is a solution betheen -2<x<-1
y=2^(-x)
y=2^(2x)+3
==>2^(2x)+3=1/2^x
==>2^(3x)+3*2^x-1=0 (1)
Let's assume u=2^x
(1)==>u^3+3*u-1=0
which as 3 roots
u=0.322185354626 or
u = -0.161092677313 + i1.754380959784 or
u = -0.161092677313 - i1.754380959784.
Let's take the real solution
0.322185354626=2^x
==>x=ln(0.322185354626) / ln(2)
x=-1,6340371790199...
an other way is
f(x)=2^(3x)+3*2^x-1
f(-2)=1/64+3/4-1=-15/64 <0
f(-1)=1/8+1-1=1/8>0
==> there is a solution betheen -2<x<-1