Equation

[DanieldeLucena]

I don't know how to desenvolve the rest of this equation to find the roots
[attachment=0:3fdc4c5h]1.JPG[/attachment:3fdc4c5h]
It's equal to Zero

Can anyone show me the next step ???

 

[mmm4444bot]

Use properties of exponents, to obtain an equation in quadratic form.

2^(2x) can be written as (2^x)^2

2^(x + 2) can be written as 2^x * 2^2

After making these substitutions, follow the Order of Operations to get:

(2^x)^2 - 20(2^x) + 64 = 0

Solve this quadratic, and check your results.

Please show us how far you get, and we'll go from there.

If you get confused, it will help to ask us specific questions.

 

[soroban]

Hello, DanieldeLucena!

\(\displaystyle \text{Solve for }x\!:\;\;2^{2x} - 5\cdot2^{x+2} + 64 \:=\:0\)

\(\displaystyle \text{We have: }\;2^{2x} - 5\!\cdot\!2^2\!\cdot\!2^x + 64 \;=\;0 \quad\Rightarrow\quad 2^{2x} - 20\!\cdot\!2^x + 64 \:=\:0\)

\(\displaystyle \text{Factor: }\;(2^x - 4)(2^x-16) \;=\;0\)

\(\displaystyle \text{And we have two equations to solve:}\)

. . \(\displaystyle \begin{array}{ccccccccccc} 2^x - 4 \;=\; 0 & \Rightarrow & 2^x \;=\; 2^2 & \Rightarrow & \boxed{x \;=\;2} \\ \\[-3mm] 2^x-16 \;=\;0 & \Rightarrow & 2^x \;=\;2^4 & \Rightarrow & \boxed{x \:=\:4} \end{array}\)

 

[mmm4444bot]

Hello, DanieldeLucena!

\(\displaystyle \text{Solve for }x\!:\;\;2^{2x} - 5\cdot2^{x+2} + 64 \:=\:0\)

\(\displaystyle \text{We have: }\;2^{2x} - 5\!\cdot\!2^2\!\cdot\!2^x + 64 \;=\;0 \quad\Rightarrow\quad 2^{2x} - 20\!\cdot\!2^x + 64 \:=\:0\)

\(\displaystyle \text{Factor: }\;(2^x - 4)(2^x-16) \;=\;0\)

\(\displaystyle \text{And we have two equations to solve:}\)

. . \(\displaystyle \begin{array}{ccccccccccc} 2^x - 4 \;=\; 0 & \Rightarrow & 2^x \;=\; 2^2 & \Rightarrow & \boxed{x \;=\;2} \\ \\[-3mm] 2^x-16 \;=\;0 & \Rightarrow & 2^x \;=\;2^4 & \Rightarrow & \boxed{x \:=\:4} \end{array}\)

Or, you can simply wait for the showoff to provide camera-ready copy. :roll:

 

[Mrspi]

I don't know how to desenvolve the rest of this equation to find the roots
[attachment=0:ne2320vi]1.JPG[/attachment:ne2320vi]
It's equal to Zero

Can anyone show me the next step ???

I wonder if a substitution might show that this equation is in "quadratic form."

First, let's note that 2[sup:ne2320vi]2x[/sup:ne2320vi] can be written as (2[sup:ne2320vi]x[/sup:ne2320vi])[sup:ne2320vi]2[/sup:ne2320vi]

And, 2[sup:ne2320vi]x + 2[/sup:ne2320vi] can be written as 2[sup:ne2320vi]2[/sup:ne2320vi]*2[sup:ne2320vi]x[/sup:ne2320vi]

Since I see 2[sup:ne2320vi]x[/sup:ne2320vi] in both the first and second terms, maybe we should do this:

Let u = 2[sup:ne2320vi]x[/sup:ne2320vi]

Make that substitution, and you'll have

u[sup:ne2320vi]2[/sup:ne2320vi] - 5*2[sup:ne2320vi]2[/sup:ne2320vi]*u + 64 = 0

Or,

u[sup:ne2320vi]2[/sup:ne2320vi] - 20u + 64 = 0

Now....can you solve that for u? And when you have the value(s) for u, remember that u = 2[sup:ne2320vi]x[/sup:ne2320vi]

Does that help?

 

[DanieldeLucena]

Yes, thx ...

 

[lookagain]

I don't know how to desenvolve the rest of this equation to find the roots

\(\displaystyle 2^{2x} - 5\times 2^{x + 2} + 64 = 0\)

It's equal to Zero

Can anyone show me the next step ???

DanieldeLucena,

I don't know if you created this form of the equation, or if it was supplied to you.

It is not good form to use a times sign with the algebraic variables.

Instead, whoever created the problem should type it as:


\(\displaystyle 2^{2x} - 5(2^{x + 2}) + 64 = 0 \ \ or\)

\(\displaystyle 2^{2x} - (5)2^{x + 2} + 64 = 0\)


***EDIT***

Thank you, Subhotosh Khan, in the post following this.
I removed the first version of the equation.

 

[Deleted member 4993]

\(\displaystyle 2^{2x} - 5\cdot2^{x + 2} + 64 = 0, \ \ or\)

This form - of using . for multiplication should be avoided - as it can easily be confused with decimal point.