Answer:
The value of x is 3 or 2.
Explanation:
Given equation is 2^(x-2) + 2^(3-x) = 3
On applying exponents rule a^(x-y) = a^x÷a^y
Where, a = 2, x = x and y = -2
So, using this, we get
⇛{(2^x/2²) + (2³/2^x)} = 3
⇛{(2^x/2*2) + {(2*2*2)/2^x}] = 3
⇛[(2^x)/4} + {(4*2)/2^x}] = 3
⇛[(2^x)/4} + {8/(2^x)}] = 3
Let assume that 2^x = y
So, above equation can be rewritten as
⇛{(y/4) + (8/y)} = 3
⇛{(y²+32)/4y} = 3
⇛{(y²+32)/4y} = (3/1)
On applying cross multiplication then
⇛1(y²+32) = 3(4y)
Multipy the number outside of the brackets with numbers and variables on the brackets on both LHS and RHS.
⇛y²+32 = 12y
⇛y²-12y + 32 = 0
By splitting the middle term, we get
⇛y² - 8y - 4y + 32 = 0
⇛y(y-8) - 4(y-8) = 0
⇛(y-8)(y-4) = 0
➝ y = 8 or y = 4
➝ 2^x = 8 or 2^x = 4
➝ 2^x = 2³ or 2^x = 2²
Therefore, x = 3 or x = 2
Answer: The value of x is 3 or 2.
VERIFICATION:
•If x = 3 the equation is
2^(x-2) + 2^(3-x) = 3
Substitute the value of x = 3 in equation
⇛2^(3-2) + 2^(3-3) = 3
⇛2¹ + 2⁰ = 3
⇛2 + 1 = 3
⇛3 = 3
LHS = RHS
•If x = 2 then the equation is
2^(x-2) + 2^(3-x) = 3
Substitute the value of x = 2 in equation
⇛2^(2-2) + 2^(3-2) = 3
⇛2^0 + 2^1 = 3
⇛1 + 2 = 3
⇛3 = 3
LHS = RHS
Hence, verified.
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