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

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

Let's start with setting the [tex]y[/tex]'s equal to find those [tex]x[/tex]'s for which the [tex]y[/tex]'s are the same.

[tex]\log(x)=\frac{1}{2}\log(x+1)[/tex]

By power rule:

[tex]\log(x)=\log((x+1)^\frac{1}{2})[/tex]

Since [tex]\log(u)=\log(v)[/tex] implies [tex]u=v[/tex]:

[tex]x=(x+1)^\frac{1}{2}[/tex]

Squaring both sides to get rid of the fraction exponent:

[tex]x^2=x+1[/tex]

This is a quadratic equation.

Subtract [tex](x+1)[/tex] on both sides:

[tex]x^2-(x+1)=0[/tex]

[tex]x^2-x-1=0[/tex]

Comparing this to [tex]ax^2+bx+c=0[/tex] we see the following:

[tex]a=1[/tex]

[tex]b=-1[/tex]

[tex]c=-1[/tex]

Let's plug them into the quadratic formula:

[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

[tex]x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}[/tex]

[tex]x=\frac{1 \pm \sqrt{1+4}}{2}[/tex]

[tex]x=\frac{1 \pm \sqrt{5}}{2}[/tex]

So we have the solutions to the quadratic equation are:

[tex]x=\frac{1+\sqrt{5}}{2}[/tex] or [tex]x=\frac{1-\sqrt{5}}{2}[/tex].

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

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

So the [tex]x[/tex] where the equations intersect is at [tex]x=\frac{1+\sqrt{5}}{2}[/tex].

Let's find the [tex]y[/tex]-coordinate.

You may use either equation.

I choose [tex]y=\log(x)[/tex].

[tex]y=\log(\frac{1+\sqrt{5}}{2})[/tex]

The intersection is [tex](\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})[/tex].