logarithmic inequality: log_0.5(2x - 1) - log_2(5 - 3x) <= 0

[Steven G]

No, those highlighted inequalities of yours in the quote box
are not equivalent to the original inequality. Yours have a different
solution set. Even if you were aware of that, you failed to
make sure the OP understood why what would later turn out
to be along my approach to be the correct way.

Information was lost. It's sort of analogous to a student
having to solve 1 - x = 1 - x.

Imagine then that a helper comes along and writes

1/(1 - x) = 1/(1 - x)

And the helper says the solution is "x = all real numbers except 1."

These were quantities that were placed in the denominator, and
part of the solution was lost.

I got the same solution as you got. I think it was (2/3,3/2).

Where you are losing me is when you say I put some quantities in the denominator. I am not saying these two are equal but let's look at them anyways. log(a) vs l0g(1/a). In the left expression a must be greater than zero, ie a>0. On the rhs, 1/a must be greater than zero, ie a>0. Nothing was lost.

Here is how I did the problem.
log_1/2(2x-1) - log₂(5-3x)<=0. The domain is x>1/2 and x<5/3. Combing yields 1/2 < x < 5/3.

log₂(1/(2x-1)) - log₂(5-3x)<=0

-log₂(2x-1) - log₂(5-3x)<=0

log₂(2x-1) + log₂(5-3x) >=0

log₂(2x-1)(5-3x) >=0

(2x-1)(5-3x)>=1 To same time, this yields 2/3 <= x <= 3/2

Now the answer is the intersection of 2/3 <= x <= 3/2 and 1/2 < x < 5/3 which is 2/3 <= x <= 3/2.

What is wrong with this? If you think it is correct and just think that I did not emphasize the domain found at the beginning I'll agree with that.

 

[tkhunny]

No disagreement from me on that either.

But I am now very curious about how best to name the fact that if f(x) is monotonic over its entire domain, then f(b) = f(c) entails that b = c.

What we can do with the log function over the entire domain of positive numbers cannot be done with the sine function.

Maybe a subset of the more general "Reversibility"?
Maybe just Injective (1-1)?