Waste less time on Facebook — follow Brilliant.

Exponential Equation

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

Note by D K
1 year ago

No vote yet
1 vote


Sort by:

Top Newest

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

Log in to reply

@Mark C Great observation​! Calvin Lin Staff · 1 year ago

Log in to reply

@Calvin Lin Thanks! It was actually not quite correct as I originally stated it, but I think it's better now. Mark C · 1 year ago

Log in to reply

@Mark C The easy way to find the minimum is to apply the arithmetic mean - geometric mean inequality.

\[ 5^ x + 5 ^ {-x} \geq 2 \times \sqrt{ 5 ^ x \times 5 ^ {-x} } = 2 \]

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

Log in to reply

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

Log in to reply

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

Log in to reply

@Paul Ryan Longhas 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. Calvin Lin Staff · 1 year ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...