lilykhan566
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Sagot :

LammettHash

Expand the equation a bit to get

[tex]2^{2x} - 3\left(2^{x+2}\right) + 32 = 0[/tex]

[tex]2^{2x} - 3\left(2^x \cdot 2^2\right) + 32 = 0[/tex]

[tex]2^{2x} - 12 \left(2^x\right) + 32 = 0[/tex]

[tex]\left(2^x\right)^2 - 12 \left(2^x\right) + 32 = 0[/tex]

Substitute y = 2ˣ and you'll notice this is really a quadratic equation in disguise. You end up with

[tex]y^2 - 12y + 32 = 0[/tex]

which is easily solved by factorizing,

[tex](y - 8) (y - 4) = 0[/tex]

[tex]\implies y - 8 = 0 \text{ or }y - 4 = 0[/tex]

[tex]\implies y = 8 \text{ or }y = 4[/tex]

Now, 2³ = 8 and 2² = 4, so

• if y = 2ˣ = 8, then x = 3; otherwise,

• if y = 2ˣ = 4, then x = 2

felipesierra213

Answer:

x = 3 or x = 2

Step-by-step explanation:

answer is in the picture

View image felipesierra213
View image felipesierra213
View image felipesierra213
View image felipesierra213
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