Solving for x: 2^(-2x+2) - 65*2^(-x) + 16 = 0

[Math wiz ya rite 09]

Solve for X:

2^(-2x+2) - 65*2^(-x) + 16 = 0

If someone could go over this step by step I would appreciate it.

Thanks!

 

[Denis]

First, can you solve this:

3(x + 2)^2 - 5(x + 2) - 112 = 0

 

[Math wiz ya rite 09]

First, can you solve this:

3(x + 2)^2 - 5(x + 2) - 112 = 0

Do you make (x+2) = y and then factor and plug it back in? I'm not sure. Could you show me, please?

Thank you.

 

[Denis]

Yes; so do it.

 

[soroban]

Hello, Math wiz ya rite 09!

There are a number of approaches . . . Here's one of them.

Solve for \(\displaystyle x:\;\;2^{-2x+2}\, -\, 65\cdot2^{-x}\,+\,16\;=\;0\)

We have: \(\displaystyle \:2^{-2x}\cdot2^2 \,-\,65\cdot2^{-x}\,+\,16 \;=\;0\)

. . . . . . \(\displaystyle 4\cdot\left(2^{-x}\right)^2 \,-\,65\left(2^{-x}\right) \,+\,16\;=\;0\)


Multiply by \(\displaystyle \left(2^{-x}\right)^2:\)

. . \(\displaystyle 4\,-\,65\left(2^x\right) \,+\,16\left(2^x\right)^2\;=\;0\)


We have: \(\displaystyle \:16\left(2^x)^2\,-\,65\left(2^x)\,+\,4\;=\;0\)

. . which factors: \(\displaystyle \:\left(2^x\,-\,4\right)\left(16\cdot2^x\,-\,1\right)\;=\;0\)


Hence: \(\displaystyle \:2^x\,-\,4\:=\:0\;\;\Rightarrow\;\;2^x\:=\:4\;\;\Rightarrow\;\;2^x\:=\:2^2\;\;\Rightarrow\;\;\fbox{x\,=\,2}\)

. .and: \(\displaystyle \:16\cdot2^x\,-\,1\:=\:0\;\;\Rightarrow\;\;2^x\:=\:\frac{1}{16}\;\;\Rightarrow\;\;2^x\:=\:2^{-4}\;\;\Rightarrow\;\;\fbox{x\:=\:-4}\)