Solve the equation:

\[ 4^x - 10 \cdot 2^x + 16 = 0 \]

4 x − 10 ⋅ 2 x + 16 = 0 ⇒ ( 2 2 ) x − 10 ⋅ 2 x + 16 = 0 ⇒ ( 2 x ) 2 − 10 ⋅ 2 x + 16 = 0
0\\\\t^2-10t+16=0\\t^2-2t-8t+16=0\\t(t-2)-8(t-2)=0\\(t-2)(t-8)=0\iff t-2=0\ or\ t-8=0\\\\t=2\ or\ t=8\\\\t=2^1\ or\ t=2^3\\\\therefore\ 2^x=2^1\ or\ 2^x=2^3\\\\Answer:\boxed{x=1\ or\ x=3}"> s u b t i t u t e 2 x = t > 0 t 2 − 10 t + 16 = 0 t 2 − 2 t − 8 t + 16 = 0 t ( t − 2 ) − 8 ( t − 2 ) = 0 ( t − 2 ) ( t − 8 ) = 0 ⟺ t − 2 = 0 or t − 8 = 0 t = 2 or t = 8 t = 2 1 or t = 2 3 t h ere f ore 2 x = 2 1 or 2 x = 2 3 A n s w er : x = 1 or x = 3 ​

Answered by Anonymous | 2024-06-10

2^x=a 4^x=(2^2)^x=(2^x)^2=a^2
a^2-10a+16=0 Δ=10^2-4 1 16=100-64=36 √Δ=6 a1=(10+6)/2=16/2=8 a2=(10-6)/2=4/2=2 2^x=8=2^3⇒ x=3 or 2^x=2=2^1⇒ x=1
There are 2 solutions : x=1 or x=3

Answered by dannielle | 2024-06-10

The solutions to the equation 4 x − 10 ⋅ 2 x + 16 = 0 are x = 1 and x = 3 . We solved the equation by rewriting it in terms of 2 x . The process involved substituting and factoring a quadratic equation.
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Answered by Anonymous | 2025-01-19