How do you solve $[(2^{2 x}) - 3] \cdot [(2^{x + 2}) + 32] = 0$ $[(2^(2x)) - 3] * [(2^(x+2)) + 32] = 0$ ?

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How do you solve $[(2^{2 x}) - 3] \cdot [(2^{x + 2}) + 32] = 0$ $[(2^(2x)) - 3] * [(2^(x+2)) + 32] = 0$ ?

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Answer 1 · 2018-04-11T14:36:39

$x \approx .7925$ $x ~~ .7925$

Explanation:

We can start by dividing one term from both side of the equation and simplifiying:

$[(2^{2 x} - 3)] \cdot [(2^{x + 2}) + 32] = 0$ $[(2^(2x)-3)]*[(2^(x+2))+32] = 0$

$[(2^{2 x} - 3)] \cdot \frac{(2^{x + 2}) + 32}{(2^{x + 2}) + 32} = \frac{0}{(2^{x + 2}) + 32}$ $[(2^(2x)-3)]*[(2^(x+2))+32]/[(2^(x+2))+32] = 0/[(2^(x+2))+32]$

$2^{2 x} - 3 = 0$ $2^(2x)-3 = 0$

$2^{2 x} = 3$ $2^(2x) = 3$

Now we can take the Log of the equation with the base being that of the number with the variable in the exponent:

${log}_{2} 2^{2 x} = {log}_{2} 3$ $log_2 2^(2x)=log_2 3$

As you may remember, this is equivalent to:

$2x*(1) = 1.585...$

Finally, solve for $x$

$x ~~ .7925$

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How do you solve $[(2^{2 x}) - 3] \cdot [(2^{x + 2}) + 32] = 0$ $[(2^(2x)) - 3] * [(2^(x+2)) + 32] = 0$ ?

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Explanation:

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How do you solve [(2^(2x)) - 3] * [(2^(x+2)) + 32] = 0[(22x)−3]⋅[(2x+2)+32]=0[(2^(2x)) - 3] * [(2^(x+2)) + 32] = 0?

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How do you solve $[(2^{2 x}) - 3] \cdot [(2^{x + 2}) + 32] = 0$ $[(2^(2x)) - 3] * [(2^(x+2)) + 32] = 0$ ?