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lemon14 · Mar 21, 2011, 01:52 PM

$log_{\frac{3}{10}} (\frac{x}{3})= log_{\frac{3}{5}} (6-6x)$

ebaines · Mar 21, 2011, 02:35 PM

First note that because logarithms are defined only for numbers > 0 from the left hand side you know x >0, and from the right hand side you know that x <1. Now consider how you could make the left hand side = 1. If you use that value for x in the right hand side, what do you get?

lemon14 · Mar 22, 2011, 10:52 AM

According to what you said I got x=9/10, which is the correct answer, but I don't understand why to make the left hand side = 1. This is the only method that can be applied?

lemon14 · Mar 22, 2011, 11:05 AM

And what if the roots are > 1 ? Like in this case: $log_2 (x-5) = log_5 (2x+7)$

ebaines · Mar 22, 2011, 11:37 AM

I can't find a closed form solution that you can solve. But one line of reasoning that helps lead to this solution goes like this:

1. Let a = 3/10. The formula you're tryng to solve is:

$ \log_a (\frac x 3) = \log_ {2a} (6 - 6x) $

2. Apply the conversion:
$ \log _{2a} b = \frac {\log_a b} {\log _ a 2a}: \\ \\ \log _a (\frac x 3 ) = \frac 1 {\log _a 2a} \cdot \log _a (6-6x) = 1.6259 \log _a (6-6x)$

3. Use the fact that if
$ log _a b = c$ then $a^c = b$ :

$ \frac x 3 = (6-6x)^{1.6259} $

4. You know that the left hand side is less than 1/3, since x < 1. For the right hand side to be less than 1/3 means that:

$ (6-6x)^1.625<\frac 1 3 \\ 6 - 6x < (\frac 1 3)^{0.424 } = 0.627\\ 1 - \frac 1 6 (\frac 1 3)^{0.424 } <x \\ 0.895 < x $

5. So now you know that 0.895 < x < 1.

At this point you can see that x = 9/10 makes for a simple answer. The alternative is to use a numerical technique to estimate the value of x. Usoing something like Newton's method zeroes in on x = 9/10 very quickly.

ebaines · Mar 22, 2011, 11:47 AM

Originally Posted by lemon14

And what if the roots are > 1 ? Like in this case?

For $\log _2 (x-5) = \log_5 (2x +7)$ just consider numbers that would work nicely. It'll take some trial and error. For example, in considering log base 5 the kinds of numbers that work "nicely" are 5, 5^2, 5^3, etc. Try them out and see what you get. For example, if the right hand side is set to $\log _5 (5) = 1$ , then 2x+7 = 5 and x = -1. Now see if x = -1 satisfies the equation. Clearly it doesn't. So try the value of x that gives the next "nice" result for the right hand side of $log _ 5 (25)$ and see what happens.