find the exact solution of equation?

Question

(8^ x )[(1/4)^x+1] =( 2^-x ) (16)

Answer 1

let y = 2^x
y^3( 1/y^2 + 1 ) = 16/y
y^2 + y^4 = 16
(y^2+1/2)^2 = 16 + 1/4 = 65/4
y^2 + 1/2 = sqrt(65)/2
[ignore -ve root if looking for real x]
y = sqrt( (sqrt(65)-1)/2 )
[ ignore -ve root again if looking for real x]
x = ln( sqrt( (sqrt(65)-1)/2) ) / ln 2
= ln( sqrt(65)-1 ) / ln2 -1

Answer 2

I would start by rewriting each term as an exponential power of 2.
8^x would become (2^3)^x or 2^(3x)
(1/4)^(x + 1) would become (2^-2)^(x + 1) or 2^(-2x - 2)
2^-x stays like it is
16 would be written as 2^4.

Now you can combine some terms (add the exponents) on both sides.
On the left: 2^(x - 2)
On the right: 2^(4 - x)

Now, since you have the same base on both sides, you can equate their exponents to get: x - 2 = 4 - x
Solve this equation: 2x = 6 or x = 3.

Hope that helps!

Answer 3

Lets call 2^x= z
so 8^x =z^3 and (1/4)^x=1/z^2
so
z^3(1/z^2+1) =16/z
z+z^3 = 16/z so z^4 +z^2 -16 = 0
z^2=(-1+-sqrt(65) /2 Take only the + sign as Z^2>0
z^2=3.5311 and z= 1.8791 (only+ as 2^x is always positive)
so 2^x= 1.8791 and x= log(1.8791)/log2=
0.9101

Answer 4

The reason there are different answers here, is because your question has been interpreted in different ways. Should the inside of your square brackets be ((1/4)^x) + 1 or should it be
(1/4)^(x + 1).
You really need to make sure with brackets that what you are writing can only be read in one way.

Answer 5

first, put everything in terms of 2^x
(2^3x)[2^(-2x-2)]=(2^-x)(2^4)
combine the terms
2^(x-2)=2^(4-x)
x-2=4-x
x=6-x
0=6-2x
2x=6
x=3