Question

Intersection point of Y=logx and y=1/2log(x+1)

Answer 1

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$.

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$, we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$'s equal to find those $x$'s for which the $y$'s are the same.

$\log(x)=\frac{1}{2}\log(x+1)$

By power rule:

$\log(x)=\log((x+1)^\frac{1}{2})$

Since $\log(u)=\log(v)$ implies $u=v$:

$x=(x+1)^\frac{1}{2}$

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

$b=-1$

$c=-1$

Let's plug them into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

$x=\frac{1+\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}$.

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$.

Let's find the $y$-coordinate.

You may use either equation.

I choose $y=\log(x)$.

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$.