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# How do you solve ${4^{{x^2} + 4x}} = {2^{ - 6}}$?

Last updated date: 09th Aug 2024
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Answer
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Hint: Here in this question, we have to solve this question. The given question is in the form of an exponential number. It is defined as the number of times the number is multiplied by itself. By using the definition of exponential number and the law of indices we are solving the given question.

Complete step-by-step solution:
The exponential number is defined as the number of times the number is multiplied by itself. Here we have to find the value of x. Consider the given equation
${4^{{x^2} + 4x}} = {2^{ - 6}}$------- (1)
Here in the above equation the first term present in LHS of the equation are the multiples of 2.
The exponential form of 4 is written as ${2^2}$ ---- (2)
Substitute the equation (2) in the equation (1). So the given equation is rewritten as
$\Rightarrow {2^{2({x^2} + 4x)}} = {2^{ - 6}}$
Hence by simplifying the exponents of the above equation.
$\Rightarrow {2^{2{x^2} + 8x}} = {2^{ - 6}}$
According to the properties of exponential numbers, if the value of the base is the same then we can equate the exponents. So we can write the above equation as
$\Rightarrow 2{x^2} + 8x = - 6$
Take -6 to the LHS, the equation can be written as
$\Rightarrow 2{x^2} + 8x + 6 = 0$
Divide the above equation by 2 we have
$\Rightarrow {x^2} + 4x + 3 = 0$
The above equation is written as
$\Rightarrow {x^2} + 3x + x + 3 = 0$
Take x as common from first two term and 1 as common from last terms so we have
$\Rightarrow x(x + 3) + 1(x + 3) = 0$
Take (x+3) as common in the above equation we have
$\Rightarrow (x + 3)(x + 1) = 0$
On simplification we have
$\Rightarrow (x + 3) = 0$ and $(x + 1) = 0$
Hence we have
$\Rightarrow x = - 3$ and $x = - 1$
Therefore, we have solved the given question.
Therefore $x = - 3$ and $x = - 1$

Note: The exponential number is an inverse of the logarithmic function. To solve we can apply the log on both sides but here we have used the definition of the exponential number we convert the number to the exponential number. The law of indices is used to solve these kinds of problems.