ch indices ...please answer fast i have exam tomorrow
2^2x+2^x+2-4*2^3=0
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2^2x + 2^(x+2) – 4*2³ = 0
(2^x)²+2^x×2² – 4×8 = 0
(2^x)² + 4 × 2^x – 32 = 0
(2^x)² + 8 × 2^x – 4×2^x – 32 = 0
(2^x+8)×(2^x–4)=
2^x = 4 or –8
2^x = 2²
x = 2
(2^x)²+2^x×2² – 4×8 = 0
(2^x)² + 4 × 2^x – 32 = 0
(2^x)² + 8 × 2^x – 4×2^x – 32 = 0
(2^x+8)×(2^x–4)=
2^x = 4 or –8
2^x = 2²
x = 2
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dhirendra1419:
but the answer given in my book is 2
upload question snap once
from text book
its 2^x+2.....2 is also a power
2^(2x)+2^(x+2)-4*2^3=0
2^2x + 2^(x+2) – 4*2³ = 0
(2^x)²+2^x×2² – 4×8 = 0
(2^x)² + 4 × 2^x – 32 = 0
(2^x)² + 8 × 2^x – 4×2^x – 32 = 0
(2^x+8)×(2^x–4)=
2^x = 4 or –8
2^x = 2²
x = 2
(2^x)²+2^x×2² – 4×8 = 0
(2^x)² + 4 × 2^x – 32 = 0
(2^x)² + 8 × 2^x – 4×2^x – 32 = 0
(2^x+8)×(2^x–4)=
2^x = 4 or –8
2^x = 2²
x = 2
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