The number of real solutions $cos (e^{x}) = 2^{x} + 2^{- x}$ $cos(e^x)=2^x + 2^-x$ is ?

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Answer 1 · 2018-02-25T16:59:44

See below.

$2^{x} + 2^{- x}$ $2^x+2^-x$ is an even function with minimum at $x = 0$ $x = 0$

or $2 \leq 2^{x} + 2^{- x}$ $2 le 2^x+2^-x$

also we know that

$- 1 \leq cos (y) \leq 1$ $-1 le cos(y) le 1$

and the conclusion is that

$cos (y)$ $cos(y)$ and $2^{x} + 2^{- x}$ $2^x+ 2^-x$ does not intercept so no real solution is possible for

$cos (e^{x}) = 2^{x} + 2^{- x}$ $cos(e^x)=2^x + 2^-x$

The number of real solutions cos(e^x)=2^x + 2^-xcos(ex)=2x+2−xcos(e^x)=2^x + 2^-x is ?