Difference between revisions of "2021 JMPSC Sprint Problems/Problem 12"
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==Solution 4== | ==Solution 4== | ||
− | 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) = 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>. |
+ | |||
+ | ~hh99754539 | ||
+ | |||
+ | ==See also== | ||
+ | #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]] | ||
+ | #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Latest revision as of 20:02, 12 July 2021
Problem
The solution to the equation can be written as
, where
is a real number. What is
?
Solution
Let Then, we have that the expression on the left hand side is equivalent to
Thus, we have that
Taking the 7th root of both sides gives
thus we have
which makes
Answer is
~Lamboreghini
Solution 2
Note that . So
. Simplifying gives that
. If
is
, then
, so
.
Solution 3
We square both sides of the equation to get We square both sides of the equation again to get
Thus,
, so the answer is
.
~tigerzhang
Solution 4
We can divide both sides by to get
. Squaring both sides gives
. Dividing both sides by
gives
. Squaring both sides again gives
. Dividing both sides gives
. We can factor this as
. We know that since
is a real number,
also must be real, and since
is real,
must be real. We can take the 7th root on both sides to get
. Multiplying both sides by
gives
. We know that
, which means that
.
~hh99754539
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.