Situation
I have the following equation:
$|\log_{10}x| + (x-1)^2 = a$
Or, I think it will be better to write it in the following form:
$|\log_{10}x| = – (x-1)^2 + a$.
I think with this, we can draw the parabola way better.
The question
Find out the x when a = 82
and $x \in Z.$
What I've tried
- Well… I thought maybe we can solve this like a usual $|f(x)|=g(x)$ and find out what happens:
$|\log_{10}x|=-(x-1)^2+82$
$
\begin{cases}
-(x-1)^2+82 \geqslant 0 \\
\log_{10}x = -(x-1)^2+82 \quad OR \quad \log_{10}x=(x-1)^2-82
\end{cases}
$
After simplifying, I found out that I can't do really much here.
- Second option was to draw the graph. After doing it manually by hand(which wasn't that hard), I tried to plot it with a program (called Desmos) to see if I was missing something.
What I have found out is that
- When my a is
0
, then I have only onex
. - When my a is greater than
0
, then I have 2x
s. - When my a is less than
0
, then I have no solution.
But I wasn't able to find the intersection point.
How do you solve this kind of problems?
Best Answer
The RHS is an integer, so the LHS must also be an integer, i.e. $x$ is a power of ten. Eyeballing shows that $x=10$ works.