basic algebraic manipulation task: use p = 2^x to rewrite 2*4^x + 2^{x+3} = 1 + 2^{x-2} as 8p^2 + 31p - 4 = 0

[noor ul hassan]

Was struggling to attempt a question on disguised quadratics, here is the question. I will list my proposed solution underneath.

By letting [imath]p = 2^x[/imath], show that the equation:

[imath]\qquad 2 * 4^{x} + 2^{x+3} = 1 + 2^{x-2}[/imath]

can be rewritten in the form:

[imath]\qquad 8p^2 + 31p - 4 = 0[/imath]

I have rewritten the exponential term with base 4 as [imath]4^x = (2^x)^2 = p^2[/imath], and have considered writing [imath]2^{x+3} [/imath] as [imath]2^{x} * 2^{3}[/imath] which equals [imath]8p[/imath]. Likewise I have manipulated the [imath]2^{x-2}[/imath] to be [imath]2^x / 2^2[/imath] which can be written as [imath]4^{-1}p[/imath].

After collecting my terms, I still failed to reach the final form [imath]8p^{2} + 31p - 4 = 0[/imath].

This thread was partially me wanting to use LaTeX to write satisfying exponential equations, but also me being slow with regards to basic arithmetic. Either way, I would love for somebody to nudge me in the right direction.

Thanks

 

[skeeter]

start by multiplying every term by 4 ...

[imath]8 \cdot 2^{2x} + 32 \cdot 2^x = 4 + 2^x[/imath]

 

[Steven G]

2^x+3= 2^x2³= 2^x*8 = 8*2^x​

2^x-2= 2^x/2²=2^x//4​

4^x=[2²]^x=(2^x)²​