How do I find a point on a graph which is equal on both the axis? I have the equation $ 10^{x-0.7711} = x $. 
In order to find x, I thought that I'll graph the equation, and the point where x = y, will be the answer. 
How do I do this? Or is there any other way to solve this equation?
 A: For sure, graphing the function is a good idea since it will show where are the roots ... if they exist !
Let us consider for more generality function $$f(x)=10^{x-a}-x$$ Its derivative is $$f'(x)=10^{x-a}\,\log (10) -1$$ the point where it cancels is given by $$x_*=a-\frac{\log (\log (10))}{\log (10)}$$ At this point $$f(x_*)=-a+\frac{\log (\log (10))}{\log (10)}+\frac{1}{\log (10)}$$ The second derivative $$f''(x)=10^{x-a}\,\log ^2(10) $$ being always positive, then $x_*$ corresponds to a minimum ($f''(x_*)=\log(10)$).
So, in order to have roots, we must have $$f(x_*)=-a+\frac{\log (\log (10))}{\log (10)}+\frac{1}{\log (10)}\leq 0$$ which means that $$a \geq\frac{\log (\log (10))}{\log (10)}+\frac{1}{\log (10)}\approx 0.79651$$
So, for the case where $a<0.79651$, the equation does not show any root. If $a=0.79651$ there will be a double root and if $a>0.79651$ there will be two  roots to the equation. 
After this preliminary analysis, if there are roots, the graph of the function will show where they are approximately. Let us consider the case where $a=0.8765$. A plot of the function reveals two roots respectively close to $0.2$ and $0.8$.
To get more accurate solutions, you can use Newton method which, starting from a guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Let us apply to the first root; then we have $$x_0=0.200000$$ $$x_1=0.220621$$ $$x_2=0.221111$$ $$x_3=0.221112$$ which is the solution for six significant figures.
Let us apply to the second root; then we have $$x_0=0.800000$$ $$x_1=0.75864$$ $$x_2=0.753762$$ $$x_3=0.753697$$ $$x_4=0.753696$$ which is the solution for six significant figures.
