Math Olympiad | A Nice Algebra Problem | A Nice Exponential Equation

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Find the value of x?
How to solve 16^x=x^2
In this video, we'll show you How to Solve Math Olympiad Question A Nice Exponential Equation 16^x=x^2 in a clear , fast and easy way. Whether you are a student learning basics or a professtional looking to improve your skills, this video is for you. By the end of this video, you'll have a solid understanding of how to solve math olympiad exponential equations and be able to apply these skills to a variety of problems.
#matholympiad #mathtricks #maths

[@2012tulio]

-1/2 is the real solution but there is also an imaginary number for this equation

[@SALogics]

You are right, I appreciate that ❤

[@DedMatveev]

There are 2 mistakes: 1) (x^2)^1/2=|x| 2) 4^x=x => x>0 and x can't be equal to -1/2. Fortunately for the author, these mistakes cancelled each other out.

[@SALogics]

Thanks for your feedback ❤

[@payoo_2674]

Use the Lambert W function W(■*e^■) = ■ 16^x = x^2 ln(16^x) = ln(x^2) x*ln(16) = 2*ln|x| ===> two cases 1st case: x > 0 x*ln(16) = 2*ln(x) ln(x)*x^(-1) = ln(16)/2 ln(x)*e^ln(x^(-1)) = ln(16)/2 ln(x)*e^(-ln(x)) = ln(16)/2 -ln(x)*e^(-ln(x)) = -ln(16)/2 W(-ln(x)*e^(-ln(x))) = W(-ln(16)/2) -ln(x) = W(-ln(16)/2) ln(x) = -W(-ln(16)/2) x = e^(-W(-ln(16)/2)) ===> -ln(16)/2 < -1/e ===> no real solutions in this case (but a lot of imaginary) 2nd case: x < 0 x*ln(16) = 2*ln(-x) ln(-x)*x^(-1) = ln(16)/2 -ln(-x)*x^(-1) = -ln(16)/2 ln(-x)*(-x)^(-1) = -ln(16)/2 ln(-x)*e^ln((-x)^(-1)) = -ln(16)/2 ln(-x)*e^(-ln(-x)) = -ln(16)/2 -ln(-x)*e^(-ln(-x)) = ln(16)/2 W(-ln(-x)*e^(-ln(-x))) = W(ln(16)/2) -ln(-x) = W(ln(16)/2) ln(-x) = -W(ln(16)/2) -x = e^(-W(ln(16)/2)) x = -e^(-W(ln(16)/2)) ===> ln(16)/2 > 0 ===> 1 real solution in this case (and a lot of imaginary) x = -e^(-W₀(ln(16)/2)) = -e^(-W₀((-1/4)*ln(2^4)/((-1/4)*2))) = -e^(-W₀(ln((2^4)^(-1/4))/(-1/2))) = = -e^(-W₀(-ln(1/2)/(1/2))) = -e^(-W₀(-ln(1/2)*(1/2)^(-1))) = -e^(-W₀(-ln(1/2)*e^ln((1/2)^(-1)))) = = -e^(-W₀(-ln(1/2)*e^(-ln(1/2)))) = -e^(-(-ln(1/2))) = -e^(ln(1/2)) = -1/2

[@SALogics]

Very nice trick! I really appreciate that ❤

[@AbbasGuclu-jn3po]

x=-1/2

[@SALogics]

Yes, you are right ❤

[@fabioalal]

IT S WRONG.SORRY

[@jamesharmon4994]

You're wrong, sorry. 😅

[@SALogics]

The answer is verified in the end, plz watch complete video❤