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# The equation ${x^{{x^{{x^ + }}}}} = 2$ is satisfied when $x$ is equal toA. InfinityB. 2C. $\sqrt{2}$D. $\sqrt 2$ Verified
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Hint: In the provided equation of infinite exponent power, first use the power rule, which is $\ln \left( {{x^x}} \right) = x\ln x$ nd then substitute the value from the given equation in the generated equation. Then, on each of the sides, we'll use logarithm and exponential to determine the required value.

Given that the equation is ${x^{{x^{{x^ + }}}}} = 2$.
Taking logarithm in the above equation on each of the sides, we get
$\ln \left( {{x^{{x^{{x^{{x^x}}}}}}}} \right) = \ln 2$
We know that when a logarithmic term has an exponent, the logarithm power rule tells us that we can transfer the exponent to the front of the logarithm.
We will now use the power rule, that is, $\ln \left( {{x^x}} \right) = x\ln x$, in left side of the above equation.
${x^{{x^{{x^x}}}}} \cdot \ln \left( x \right) = \ln 2$
Substituting the value of ${x^{{x^{{x^x}}}}}$ in the above equation, we get
$2 \cdot \ln \left( x \right) = \ln 2$
Dividing the above equation by 2 on each of the sides, we get
$\Rightarrow \dfrac{{2 \cdot \ln \left( x \right)}}{2} = \dfrac{{\ln 2}}{2}$
$\Rightarrow \ln x = \dfrac{{\ln 2}}{2}$
Rearranging the right side of the above equation, we get
$\Rightarrow \ln x = \dfrac{1}{2}\ln 2$
Using the power rule on the right side of the above equation, we get
$\ln \left( {{2^{\frac{1}{2}}}} \right) = \ln x$
Using the formula ${2^{\frac{1}{2}}} = \sqrt 2$ in the above equation, we get
$\ln \left( {\sqrt 2 } \right) = \ln x$
Taking exponential on each of the sides in the above equation, we get
$\Rightarrow \sqrt 2 = x$
$\Rightarrow x = \sqrt 2$
Thus, the given equation satisfies only when $x = \sqrt 2$.
Hence, the correct answer is option (D).

Note: You should be familiar with the infinite exponential power, the power law of the logarithm, and exponential functions in order to solve these types of questions. One may become perplexed by the well-known fact that ${2^{\frac{1}{2}}} = \sqrt 2$, or else the learner may become perplexed. The trick to solving this question is to use the logarithm on both sides.