Intersect From Two Logarithmic Functions

[JDK]

It's funny because I've never had any problems with Math until I got to Pure Math 30. Now I feel like an idiot and have incredible urges to scream at my book as if it were listening. Please help me out here.

The graphs of y=log(base3)(x-1) +1 and y=log(base3)(2x +1) intersect at a point. An equation that could be used to find this point of intersections is...

Multiple Choice follows the Q. But, that's not what matters. I need some help on how to figure such a question out. I wish I could show some of my work to prove I've tried, but that's the thing - I'm blank right now. I don't know where to begin. I'd imagine one would combine the equations first like this...

y = log(base3)(x-1) + log(base3)(2x +1)

...and then work from there. Erghhh... take deep breaths... that's what I keep telling myself...

 

[mathman]

Since we're talking about base 3, I won't write it.

You want to find x where

log(x-1) + 1=log(2x+1)

first 1=log(3). Using sum of logs=log of product, we get

log(3x-3)=log(2x+1)

3x-3=2x+1

x=4

 

[Hurkyl]

The graphs of y=log(base3)(x-1) +1 and y=log(base3)(2x +1) intersect at a point. An equation that could be used to find this point of intersections is...

Let's move a step at a time.

You are looking for the coordinates for the point of intersection. We should give it a name to make it easier to refer to it; let's call it P. We are looking for its coordinates, so let's select variables to represent it's coordinates; how about a and b.

(I'm using a and b instead of x and y to illuminate the fact that these are conceptually different than the x and y given in the problem)

So we're looking for the coordinates of P, which are $(a, b)$.

We know P lies on the graph of $y = \log_3 (x-1) + 1$. Thus, its coordinates must satisfy the equation $b = \log_3 (a-1) + 1$.

We also know P lies on the graph of $y = \log_3 (2x+1)$. Thus, its coordinates must satisfy the equation $b = \log_3 (2a+1)$.

So, we have this system of equations we wish to solve:

$$\begin{align} b &= \log_3 (a-1) + 1 \\ b &= \log_3 (2a+1) \end{align}$$

A common step to solve a system of equations is to solve one equation for one variable, and substitute the result in another equation. Here, both equations come pre-solved for b! So we take the solution from (1) and substitute into (2) to get

$$\log_3 (a-1) + 1 = \log_3 (2a+1)$$

Which is the point where mathman started.

 

[JDK]

Great! I get it. Something in my brain just wasn't connecting properly (hasn't been for a while for some reason). The piece I was missing was that the two equations equal each other for the fact that they both have one single pair of values which satisfy one another and vice versa. Phew. Glad you guys helped out. The life of my textbook has been saved from my rage. Thanks!

I think I'm cramming to much in my head at once. I'm currently learning CSS, Javascript and Visual Basic and all my mathematical/logical decision making is getting mixed up. Maybe if I drop a language for a while I can focus better on what counts right now.