Question

Solve for x (2^2x+4 - 6.2^2x+1) / 2^x+1 = 16

Answer 1

????????????????? dont hav NE idea

yu need tu yuze parens & brakets so sumwon kan figger out wot be the base

& wot be the power

Answer 2

Solve for x (2^2x+4 - 6.2^2x+1) / 2^x+1 = 16

I'm going to put some brackets in this expression also. To give

(2^(2x+4) - 6*2^(2x+1)) / 2^(x+1) = 16

Using 2^(2x+4) = 2^(2x)*2^4, etc. the expression becomes

(2^(2x)*2^4 - 6*2^(2x)*2^1) / (2^(x)*2^1) = 16

(16*2^(2x) - 12*2^(2x)) / (2*2^(x)) = 16

(16*2^(2x) - 12*2^(2x)) = 16*(2*2^(x))

4*2^(2x) = 32*2^(x)

2^(2x) = 8*2^(x)

Let u = 2^x, then

u^2 = 8u

u^2 - 8u = 0

u(u - 8) = 0

u = 0, u = 8

Therefore, 2^x = 0, 2^x = 8

i.e. x = -infinity, x = 3

x = -infinity is a solution of the quadratic eqn, but not of the original expression, so we ignore it.

The solution is: x = 3

Answer 3

I read this as (2^(2x+4)-6*2^(2x+1))/2^(x+1)=16

2^(2x+4)-6*2^(2x+1)=2^4*2^(x+1)=2^(x+5)

Divide through by 2^2x:

2^4-6*2=2^(5-x)

16-12=4=2^2=2^(5-x)

Equating exponents: 2=5-x, so x=3.

CHECK

(2^10-6*2^7)/2^4=(1024-6*128)/16=(1024-768)/16=256/16=16 OK