1. ## log problem

Hi; y = log(2x - 6) Don't understand why I would factor out the 2 to give me the following? y = log2(x - 3) Whats the point?Thanks.

2. ## Re: log problem

To determine the horizontal dilation (in this case it is of factor 1/2).

3. ## Re: log problem

Doesn't it tell me that when its in the bracket?Also how would you convert it to an exponenetial?Thanks.

4. ## Re: log problem

Ok I got the convertion bit. just need to know why factor it.

5. ## Re: log problem

Originally Posted by anthonye
Hi; y = log(2x - 6) Don't understand why I would factor out the 2 to give me the following? y = log2(x - 3) Whats the point?Thanks.
Points depend on context.

As ProveIt said: the factoring is relevant to dilation.

It would also be relevant in simplifying the following equation:

$log(y + 1) + log(2) = log(2x - 6) = log((2\{x - 3\}) = log(x - 3) + log(2) \implies log(y + 1) = log(x - 3) \implies y + 1 = x - 3$

$\implies y = x - 4.$

A lot of high school algebra is just giving you a set of tools that may be useful in many situations.

6. ## Re: log problem

Ok thanks but why is it relevent to dilation?

7. ## Re: log problem

Originally Posted by anthonye
Ok thanks but why is it relevent to dilation?
$Let\ u = log_w(x - 3)\ and\ u = log_w(2v - 6) \implies log_w(x - 3) = log_w(2\{v - 3\}) = log_w(v - 3) + log_w(2) \implies$

$log_w(x - 3) - log_w(v - 3) = log_w(2) \implies log_w\left(\dfrac{(x - 3)}{(v - 3)}\right) = log_w(2) \implies \dfrac{x - 3}{v - 3} = 2 \implies$

$x - 3 = 2v - 6 \implies x = 2v - 3.$

$\therefore For\ large\ v,\ \dfrac{x}{v} = \dfrac{2v - 3}{v} = 2 - \dfrac{3}{v} \approx 2.$

In English, as y gets larger and larger, the horizontal distance between y = log(x - 3) and the y-axis gets closer and closer to double the distance between y = log(2x - 6) and the y-axis. For example,

$log_{10}(x - 3) = 1 \implies x = 13.\ And\ log_{10}(2v - 6) = 1 \implies v = 8.\ And\ \dfrac{13}{8} \approx 1.625 \approx 2.$

$log_{10}(x - 3) = 2 \implies x = 103.\ And\ log_{10}(2v - 6) = 2 \implies v = 53.\ And\ \dfrac{103}{53} \approx 1.94 \approx 2.$

$log_{10}(x - 3) = 3 \implies x = 1,003.\ And\ log_{10}(2v - 6) = 3 \implies v = 503.\ And\ \dfrac{1,003}{503} \approx 1.994 \approx 2.$

So 2 is relevant to understanding the dilation even though it is approximate. Frequently, an approximation is all you need.

Now consider $x = a + 3,\ where\ a > 0, and\ b = log_y(a).$

$log_w(x - 3) = log_y(a) = b.\ But\ log_w(2x - 6) = log_w(2a + 6 - 6) = log_w(2a) = log_w(a) + \log_w(2) = b + log_w(2).$

In English the graph of log(x - 3) is always EXACTLY log(2) units below the graph of log(2x - 6). Again 2 is involved although this time it is a function of 2 and is exact rather than approximate.