Find the intersection point of a logarithmic and a quadratic functions.

Question

Situation

I have following equation:

$|\log_{10}x| + (x-1)^2 = a$ or I think it will be better to write it in 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 the x when a = 82 and $x \in Z.$

What I've tried

• Well... I thought maybe we can solve 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 plot it with a program (called Desmos) to see if I was missing something.

What I have found out is that

1. when my a is 0, then I have only one x.
2. when my a is greater than 0, then I have 2 xs.
3. 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?

Answer 1

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.