Difference between revisions of "2021 JMPSC Sprint Problems/Problem 12"

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We can divide both sides by <math>x</math> to get <math>\frac{2^{63}}{x} = \sqrt{x \sqrt{x}}</math>. Squaring both sides gives <math>\frac{2^{126}}{x^2} = x \sqrt{x}</math>. Dividing both sides by <math>x</math> gives <math>\frac{2^{126}}{x^3} = \sqrt{x}</math>. Squaring both sides again gives <math>\frac{2^{252}}{x^6} = x</math>. Dividing both sides gives <math>\frac{2^{252}}{x^7} = 1</math>. We can factor this as <math>\left(\frac{2^{36}}{x}\right)^7 = 1</math>. We know that since <math>m</math> is a real number, <math>2^m</math> also must be real, and since <math>2^m</math> is real, <math>x</math> must be real. We can take the 7th root on both sides to get <math>\frac{2^{36}}{x} = 1</math>. Multiplying both sides by <math>x</math> gives <math>2^36 = x</math>. We know that <math>2^m = 2^{36}</math>, which means that <math>m = \boxed{36}</math>.
 
We can divide both sides by <math>x</math> to get <math>\frac{2^{63}}{x} = \sqrt{x \sqrt{x}}</math>. Squaring both sides gives <math>\frac{2^{126}}{x^2} = x \sqrt{x}</math>. Dividing both sides by <math>x</math> gives <math>\frac{2^{126}}{x^3} = \sqrt{x}</math>. Squaring both sides again gives <math>\frac{2^{252}}{x^6} = x</math>. Dividing both sides gives <math>\frac{2^{252}}{x^7} = 1</math>. We can factor this as <math>\left(\frac{2^{36}}{x}\right)^7 = 1</math>. We know that since <math>m</math> is a real number, <math>2^m</math> also must be real, and since <math>2^m</math> is real, <math>x</math> must be real. We can take the 7th root on both sides to get <math>\frac{2^{36}}{x} = 1</math>. Multiplying both sides by <math>x</math> gives <math>2^36 = x</math>. We know that <math>2^m = 2^{36}</math>, which means that <math>m = \boxed{36}</math>.
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~hh99754539

Revision as of 13:09, 11 July 2021

Problem

The solution to the equation $x \sqrt{x \sqrt{x}}=2^{63}$ can be written as $2^m$, where $m$ is a real number. What is $m$?

Solution

Let $y=\sqrt[4]{x}.$ Then, we have that the expression on the left hand side is equivalent to $y^4\cdot y^2\cdot y=y^7.$ Thus, we have that $y^7=2^{63}.$ Taking the 7th root of both sides gives $y=2^9,$ thus we have $\sqrt[4]{x}=2^9,$ which makes $x=2^{36}.$ Answer is $\boxed{36}.$

~Lamboreghini

Solution 2

Note that $\sqrt{x}=x^{\frac{1}{2}}$. So ${x}\sqrt{x^{\frac{3}{2}}}=2^{63}$. Simplifying gives that $x{\cdot}x^{\frac{3}{4}}=x^{\frac{7}{4}}=2^{63}$. If $x$ is $2^m$, then $\frac{7m}{4}=63$, so $m=\boxed{36}$.

Solution 3

We square both sides of the equation to get \[x^2 \cdot x\sqrt{x}=x^3\sqrt{x}=2^{63 \cdot 2}.\] We square both sides of the equation again to get \[x^6 \cdot x=x^7=2^{63 \cdot 4}.\] Thus, $x=2^{63 \cdot 4/7}=2^{36}$, so the answer is $\boxed{36}$.

~tigerzhang

Solution 4

We can divide both sides by $x$ to get $\frac{2^{63}}{x} = \sqrt{x \sqrt{x}}$. Squaring both sides gives $\frac{2^{126}}{x^2} = x \sqrt{x}$. Dividing both sides by $x$ gives $\frac{2^{126}}{x^3} = \sqrt{x}$. Squaring both sides again gives $\frac{2^{252}}{x^6} = x$. Dividing both sides gives $\frac{2^{252}}{x^7} = 1$. We can factor this as $\left(\frac{2^{36}}{x}\right)^7 = 1$. We know that since $m$ is a real number, $2^m$ also must be real, and since $2^m$ is real, $x$ must be real. We can take the 7th root on both sides to get $\frac{2^{36}}{x} = 1$. Multiplying both sides by $x$ gives $2^36 = x$. We know that $2^m = 2^{36}$, which means that $m = \boxed{36}$.

~hh99754539