\(\sin(e^x)\), 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:

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.

When solving over the complex, there is almost always a solution. It's a result from complex analysis (Picard's theorem) that given any analytic function on the plane, the image misses at most one point.

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## Comments

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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 \(x>0\).)

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Great observation!

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

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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 \( x = 0 \).

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Go Here ,The easiest Way to Solve this Problem is by using A.M-G.M inequality

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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.

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When solving over the complex, there is almost always a solution. It's a result from complex analysis (Picard's theorem) that given any analytic function on the plane, the image misses at most one point.

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