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First of all, you should know how to use standard notation about square or power.\r
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\n" ); document.write( "\n" ); document.write( " If you are in high school level and have learn calculus then it seems
\n" ); document.write( " that you have to solve the given equation by view or basic idea.\r
\n" ); document.write( "\n" ); document.write( " Direct way: it is clear that if a, b > 0 then a^b = b^a.
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\n" ); document.write( " Hence, if x = 2, then 2^x = x^2 = 2^2.
\n" ); document.write( " So, x = 2 is a root of 2^x = x^2.\r
\n" ); document.write( "\n" ); document.write( " Formal proof using logarithm and calculus as below.\r
\n" ); document.write( "\n" ); document.write( " 2^x = x^2...(*)\r
\n" ); document.write( "\n" ); document.write( " Apply log2 on both sides of(*),we have log2 x^2 = log2 2^x,
\n" ); document.write( " so 2 log2 x = x log2 2 = x,
\n" ); document.write( " Let y = log2 x , we get 2 y = 2^y .
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\n" ); document.write( " Apply log2 on both sides,we have log2(2y) = log2 2 + log2y
\n" ); document.write( " = 1 + log2y = log2 2^y = y.
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\n" ); document.write( " Let f(y) = log2 y, since f'(y) = 1/(yln2) and f(1) = 0.
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\n" ); document.write( " By the mean value theorem for f(y) on the interval [1,y] (or [y,1]),
\n" ); document.write( " there exists z' between 0 & y such that
\n" ); document.write( " (log2y - 0)/ (y -1)= f'(z') = 1/(z' ln2)
\n" ); document.write( " So, log2y = (y-1)/ (z' ln2)
\n" ); document.write( " Hence, log2y - y + 1 = (y-1)/ (z' ln2) - (y -1) = (y-1)[-1+ 1/ (z' ln2)]
\n" ); document.write( " This shows y =1 is a root of the equation log2y - y + 1 = 0.
\n" ); document.write( " And,hence 2^y = 2y has a root y=1.
\n" ); document.write( " When y = log2 x = 1, satisfies log2y - y + 1 = 0,and hence it also
\n" ); document.write( " satisfies 2^y = 2y.
\n" ); document.write( " Recall x = 2^y, so x = 2^1 = 2 if y =1.
\n" ); document.write( " We see that x =2 is a root of the given equation 2^x = x^2.
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\n" ); document.write( " Also, f(2) = log2 2 =1.
\n" ); document.write( " By the mean value theorem for f(y) on the interval [y,2] ,
\n" ); document.write( " there exists z\" between y & 2 such that
\n" ); document.write( " (log2 y - 1)/ (y -2)= f'(z\") = 1/(z\" ln2)
\n" ); document.write( " So, log2 y -1 = (y-2)/ (z' ln2)
\n" ); document.write( " Hence, log2y - y + 1 = log2 y - 1 -y + 2 = (y-2)/ (z' ln2) -(y-2)
\n" ); document.write( " = (y-2)[-1 + 1/ (z\" ln2)]
\n" ); document.write( " Thus,y = 2 satisfies log2y - y + 1 = 0.
\n" ); document.write( " Similarly, x= 2^y = 4 is also a root of the given equation 2^x = x^2.\r
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\n" ); document.write( "\n" ); document.write( " Answewr: Two roots 2 and 4. \n" ); document.write( "