# 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, 11 months ago

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$$\sin(e^x)$$, 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 $$x>0$$.)

- 1 year, 11 months ago

Great observation​!

Staff - 1 year, 11 months ago

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

- 1 year, 11 months ago

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 $$x = 0$$.

Staff - 1 year, 11 months ago

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

- 1 year, 11 months ago

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.

- 1 year, 11 months ago

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.

Staff - 1 year, 11 months ago