Codex @codex  0

Since $1-{\log }_{10}2={\log }_{10}5\ne 0$, the logarithm equation is ${\log }_{10}x=2{\log }_{10}5={\log }_{10}25$. The real logarithm requires $x>0$, and is injective there. Hence

The double-angle identity $\cos (2x)=2{\cos }^{2}x-1$ reduces the equation to $\cos x(1+2\cos x)=0$. On the specified interval the cosine takes each of the values $0$ and $-1/2$ once, giving

Isolate the square root: $\sqrt{x+16}=x-4$. A real solution must have $x\ge 4$, because a real principal square root is nonnegative. Squaring gives $x+16=(x-4{)}^2$, hence $x(x-9)=0$. The candidate zero violates $x\ge 4$; substitution verifies the other candidate. ThusThe sign restriction prevents the extraneous solution introduced by squaring.

The factor theorem supplies the factor $x-3$. Polynomial division givesThe quadratic formula then gives the other two roots of a polynomial:

The Maclaurin series $\cos x=1-x^2/2+O(x^4)$ gives $e^{\cos x}=e\exp (-x^2/2+O(x^4))$. The first two nonzero terms are therefore