What is the number of solutions to the equation \[\large \sin(e^x) = 5^x + 5^{-x} ? \]

What is the number of solutions to the equation \[\large \sin(e^x) = 5^x + 5^{-x} ? \]

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TopNewest\(\sin(e^x)\), because it is always the sine of a real number, has a maximum of 1 (whenever \(e^x = \pi/2 + k\pi\) for some integer \(k\)), and \(5^x+5^{-x}\) has its minimum at \(x=0\), where it is \(2\), so there are no solutions.

You can find that minimum with differential calculus, or in a simpler way. The function is obviously symmetric around the \(y\) axis, so we can establish that it rises away from there as \(x\) gets larger, and by symmetry we can be sure it rises in the other direction too. For \(x>0\), the function \(5^x\) increases faster than \(5^{-x}\) decreases, so the sum of those functions increases as \(x\) does:

\[5^{-(x+k)}-5^{-x}=\frac{1}{5^{x+k}}-\frac{1}{5^x} = \frac{5^x-5^{x+k}}{5^{2x+k}} = [5^{x+k}-5^x]\cdot\frac{-1}{5^{2x+k}}\]

(And \(5^{2x+k} > 1\) for \(x>0\).) – Mark C · 1 year ago

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– Calvin Lin Staff · 1 year ago

Great observation!Log in to reply

– Mark C · 1 year ago

Thanks! It was actually not quite correct as I originally stated it, but I think it's better now.Log in to reply

arithmetic mean - geometric mean inequality.

The easy way to find the minimum is to apply the\[ 5^ x + 5 ^ {-x} \geq 2 \times \sqrt{ 5 ^ x \times 5 ^ {-x} } = 2 \]

Equality holds if \( 5 ^ x = 5 ^ {-x} \), or when \( x = 0 \). – Calvin Lin Staff · 1 year ago

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Go Here ,The easiest Way to Solve this Problem is by using A.M-G.M inequality – Sambhrant Sachan · 1 year ago

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Let y = 5^x. So, 1/y = 5^(-x). Note that y, 1/y > 0. Thus, by AM-GM inequality y + 1/y > = 2.

Clearly, -1 =< sin(e^x) =< 1. Thus, sin(e^x) = 5^x + 5^(-x) is a contradiction. Therefore, no real value of x exist in this equation.

I am currently think for nonreal value of x. – Paul Ryan Longhas · 1 year ago

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complex analysis (Picard's theorem) that given any analytic function on the plane, the image misses at most one point. – Calvin Lin Staff · 1 year ago

When solving over the complex, there is almost always a solution. It's a result fromLog in to reply