How do you solve $2^{2 x} \times 4^{4 x + 8} = 64$ $2^(2x)times4^(4x+8)=64$ ?

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Answer 1 · 2016-09-18T13:03:20

$x = - 1$ $x = - 1$

We have: $2^{2 x} \times 4^{4 x + 8} = 64$ $2^(2 x) times 4^(4 x + 8) = 64$

Let's express the numbers in terms of $2$ $2$ :

$\Rightarrow 2^{2 x} \times {(2^{2})}^{4 x + 8} = 2^{6}$ $=> 2^(2 x) times (2^2)^(4 x + 8) = 2^(6)$

Using the laws of exponents:

$\Rightarrow 2^{2 x} \times 2^{8 x + 16} = 2^{6}$ $=> 2^(2 x) times 2^(8 x + 16) = 2^(6)$

$\Rightarrow 2^{2 x + 8 x + 16} = 2^{6}$ $=> 2^(2 x + 8 x + 16) = 2^(6)$

$\Rightarrow 2^{10 x + 16} = 2^{6}$ $=> 2^(10 x + 16) = 2^(6)$

$\Rightarrow 10 x + 16 = 6$ $=> 10 x + 16 = 6$

$\Rightarrow 10 x = - 10$ $=> 10 x = - 10$

$\Rightarrow x = - 1$ $=> x = - 1$

Therefore, the solution to the equation is $x = - 1$ $x = - 1$ .

How do you solve 22x×44x+8=642^(2x)times4^(4x+8)=64?