Is there any way to find the solution set, by hand, for
$7\log_{10}x=x-2$
I know $x\ge0$ and this is equivalent to $x^7=10^{x-2}$. I am aware I can simply graph $y=x^7$ and $y=10^{x-2}$ respectively, and then find when the lines intersect. Unfortunately, this would require a graphing calculator.
Are there any techniques I can possibly use by hand to find the solution set here?
Without graphing calculator and not using Lambert function.
Consider that you look for the zero's of function $$f(x)=7\log_{10}(x)-x+2$$ for which $$f'(x)=\frac{7}{x \log (10)}-1\qquad \text{and} \qquad f''(x)=-\frac{7}{x^2 \log (10)} < 0\,\, \forall x$$ The first derivative cancels at $$x_*=\frac{7}{\log (10)} \approx 3.04\implies f(x_*)\approx 2.34$$ and this point is a maximum.
Now, by inspection, $$f\left(\frac{1}{10}\right)=-\frac{51}{10} \qquad \qquad f(1)=1 \qquad\qquad f(10)=-1$$ So, we have two roots such that $$\frac{1}{10} < x_1 < 1\qquad \qquad 1 < x_2 <10$$
Now, let us use Newton method which, starting with a guess $x_0$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ To avoid overshoot of the solution, by Darboux theorem, since $f''(x) < 0$, let us start from a value $x_0$ such that $f(x_0) <0$; notice that we have these points.
Below are reproduced the iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.100000 \\ 1 & 0.273466 \\ 2 & 0.492423 \\ 3 & 0.617296 \\ 4 & 0.638658 \\ 5 & 0.639131 \end{array} \right)$$
$$\left( \begin{array}{cc} n & x_n \\ 0 & 10.0000 \\ 1 & 8.56321 \\ 2 & 8.50933 \\ 3 & 8.50923 \end{array} \right)$$