Question: Solve: $5× 4^{x+1} - 16^x = 64$


Solution:
Given,

$5× 4^{x+1} - 16^x = 64$

$or, 5× 4^x×4^1 - (4^2)^x = 64$

$or, 20 × 4^x - (4^x)^2 = 64$

[ Let $4^x$ = a ]

$or, 20a - a² = 64$

$or, a² - 20a +64 = 0$

$or, a² - (16+4)a + 64 = 0$

$or, a² - 16a -4a +64 = 0$

$or, a(a -16) - 4(a -16) = 0$

$or, (a-4)(a-16) = 0$

Either,

$(a-4) = 0$

$or, 4^x = 4^1$

$\therefore x = 1$

Or,

$(a-16) = 0$

$or, 4^x = 16$

$or, 4^x = 4^2$

$\therefore x = 2$

Hence, x = 1 or 2.


Related Notes and Solutions:

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