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zhenek [66]
2 years ago
7

Solve the equation 8·4^x+1=1​

Mathematics
1 answer:
stepladder [879]2 years ago
4 0

Answer:

See work below

Step-by-step explanation:

8(4^x)+1=1

9(4^x)=1

-9        -9

(4^x)=-8

/-8     /-8

And this is where I'm stuck, can someone help me finish?

       

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So they need to charge at least $7.54 to make back their money plus $1500.

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Using these complex zeros (1,1,-1/2,2+i,2-i) factor f(x)=-2x^5 +11x^4 -22x^3 +14x^2 +4x -5
Marianna [84]
\bf \begin{cases}
x=1\implies &x-1=0\\
x=1\implies &x-1=0\\
x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\
x=2+i\implies &x-2-i=0\\
x=2-i\implies &x-2+i=0
\end{cases}
\\\\\\
(x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0}
\\\\\\
(x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}

\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)+1]
\\\\\\
(x^2-2x+1)(2x+1)~[x^2-4x+5]
\\\\\\
(x^2-2x+1)(2x+1)(x^2-4x+5)

of course, you can always use  (x-1)(x-1)(2x+1)(x²-4x+5)  as well.
7 0
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