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

absolute valuelogarithmsquadratics

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.

List item

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?

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.

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