2to the powwe2x-3.2to the power x+2+32=0
Answers
Answer:
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Answer:
x=3 , x=2
Step-by-step explanation:
\begin{gathered}2^{2x}-3\times \:2^{(x+2)}+32=0 \\ \\\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c\\2^{x+2}=2^x\times \:2^2\\2^{2x}-3\times \:2^x\times \:2^2+32=0\\\mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=(a^b)^c\\\\2^{2x}=(2^x)^2\\(2^x)^2-3\times \:2^x\times \:2^2+32=0\\\\\mathrm{Rewrite\:the\:equation\:with\:}2^x=u\\(u)^2-3u\times \:2^2+32=0\\\\\mathrm{Expand\:}u^2-3u\times \:2^2+32:\quad u^2-12u+32\\\end{gathered}
2
2x
−3×2
(x+2)
+32=0
Applyexponentrule:a
b+c
=a
b
a
c
2
x+2
=2
x
×2
2
2
2x
−3×2
x
×2
2
+32=0
Applyexponentrule:a
bc
=(a
b
)
c
2
2x
=(2
x
)
2
(2
x
)
2
−3×2
x
×2
2
+32=0
Rewritetheequationwith2
x
=u
(u)
2
−3u×2
2
+32=0
Expandu
2
−3u×2
2
+32:u
2
−12u+32
[/tex]\begin{gathered}u=\frac{-(-12)+\sqrt{(-12)^2-4\times \:1\times \:32}}{2\times \:1}:\quad 8\\u=\frac{-(-12)-\sqrt{(-12)^2-4\times \:1\times \:32}}{2\times \:1}:\quad 4\\\\\mathrm{The\:final\:solutions\:to\:the\:quadratic\:equation\:are:}\\u=8,\:u=4\\\mathrm{Since\:}u=2^x\mathrm{,\:solve\:the\:following\:in\:order\:to\:find\:}x\\\mathrm{Solve\:}\:2^x=8:\quad x=3\\\mathrm{Solve\:}\:2^x=4:\quad x=2\\x=3,x=2\end{gathered}
u=
2×1
−(−12)+
(−12)
2
−4×1×32
:8
u=
2×1
−(−12)−
(−12)
2
−4×1×32
:4
Thefinalsolutionstothequadraticequationare:
u=8,u=4
Sinceu=2
x
,solvethefollowinginordertofindx
Solve2
x
=8:x=3
Solve2
x
=4:x=2
x=3,x=2
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