2^{2x} - 6×2^{x+1} + 32 = 0 | Indices - Solve | SciPiPupil

Question: Solve: $2^{2x} - 6.2^{x+1} + 32 = 0$

Solution:

Given,

$2^{2x} - 6.2^{x+1} + 32 = 0$

[ Let 2^x = a ]

$or, (2^x)^2 - 6×2^x×2^1 + 32 = 0$

$or, a² - 12a +32= 0$

$or, a²-(8+4)a +32=0$

$or, a² -8a -4a + 32 = 0$

$or, a(a-8) - 4(a -8)= 0$

$or, (a-4)(a-8) = 0$

Either,

$(a-4) = 0$

$or, 2^x = 4$

$or, 2^x = 2^2$

$\therefore x = 2$

Or,

$(a-8) = 0$

$or, 2^x = 8$

$or, 2^x = 2^3$

$\therefore x = 3$

Hence, (x = 2 or 3).

Related Notes and Solutions:

Here is the website link to the notes of Indices.

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