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Gekata [30.6K]
2 years ago
7

Solve the equation 2^(2x+1)+8=17×2^x.

Mathematics
1 answer:
myrzilka [38]2 years ago
8 0
Oooh, looks fun

ok
erm
we might want to know some properties
(x^m)(x^n)=x^{m+n}
(a^b)^c=a^{bc}
if a^m=a^n where a=a, then assume m=n

2^{2x+1}=(2^{2x})(2^1)
so
(2^{2x})(2)+8=17(2^x)
(2^{2x})(2)+2^3=17(2^x)
(2^x)^2(2)+2^3=17(2^x)
minus 17*2^x both sides
(2^x)^2(2)+2^3-17(2^x)=0
use u subsitution, u=2ˣ
(u)^2(2)+2^3-17(u)=0
solve 2u^2+8-17u=0
or
2u^2-17u+8=0
ac method
2 times 8=16
what 2 number multiply to get -17 and add to get 16
-16 and -1
2u²-1u-16u+8=0
(2u²-1u)+(-16u+8)=0
u(2u-1)+(-8)(2u-1)=0
(u-8)(2u-1)+0
u-8=0
u=8

2u-1=0
2u=1
u=1/2

now
u=2ˣ

8=2ˣ
2^3=2^x
3=x

and
\frac{1}{2}=2^x
2^{-1}=2^x
-1=x

x=-1 and 3
neat problem
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The missing figure is attached with the answer.

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