Solve the following logarithmic equation over real numbers

Question

Solve the equation:

$$\log_{2020} {(x^{10} + x^9 + x^8 + x^7 + x^6+ x^5+ x^2 )}=\log_2 x$$

over real numbers.

I found out that $x=2$ is a solution and I suspect is the only one, but cannot prove it.

Answer 1

Solve for $x$ in $\log_{2020} {(x^{10} + x^9 + x^8 + x^7 + x^6+ x^5+ x^2 )}=\log_2 x$.

$$\log_2{2020} =\log_x{(x^{10} + x^9 + x^8 + x^7 + x^6+ x^5+ x^2 )}$$ $x=2$ is a solution on checking.

To check if it is the only solution, let $t=\log_2 2020$ and consider \begin{align*} f(x)&=\ln(x^{10}+x^9+\cdots+x^2)-t\ln x\\ f^\prime(x)&=\frac{(10-t)x^{10}+(9-t)x^9+\cdots+(2-t)x^2}{x(x^{10}+x^9+\cdots+x^2)}<0\ \forall\ x\in[0,+\infty)&(\because 10<t) \end{align*} Since $f(x)$ is strictly decreasing, it cuts the $x-$axis at $x=2$ only.