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

[Trzewikodziobowna]

Hello. I am trying to tackle this inequality:

. . . . .\(\displaystyle \log_{\frac{1}{2}}(2x\, -\, 1)\, -\, \log_2(5\, -\, 3x)\, \leq\, 0\)

First, find the domain for each of the logs:

. . .\(\displaystyle \log_{\frac{1}{2}}(2x\, -\, 1):\)

. . . . .\(\displaystyle 2x\, -\, 1\, >\, 0\)
. . . . .\(\displaystyle x\, >\, \dfrac{1}{2}\)

. . .\(\displaystyle \log_2(5\, -\, 3x):\)

. . . . .\(\displaystyle 5\, -\, 3x\, >\, 0\)
. . . . .\(\displaystyle x\, <\, \dfrac{5}{3}\)

Then should I get rid of the log, by changing the base of 2nd log to 1/2?

. . . . .\(\displaystyle \log_a(b)\, =\, \dfrac{\log_c(b)}{\log_c(a)}\)

. . .\(\displaystyle \log_{\frac{1}{2}}(2x\, -\, 1)\, -\, \log_2(5\, -\, 3x)\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, \log_{\frac{1}{2}}(2x\, -\, 1)\, -\, \dfrac{\log_{\frac{1}{2}}(5\, -\, 3x)}{\log_{\frac{1}{2}}(2)}\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, \log_{\frac{1}{2}}(2x\, -\, 1)\, -\, \log_{\frac{1}{2}}(5\, -\, 3x)\, -\, \log_{\frac{1}{2}}(2)\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, 2x\, -\, 2\, -\, (5\, -\, 3x\, -\, 2)\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, 2x\, -\, 1\, -\, 5\, +\, 2x\, +\, 2\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, 5x\, -\, 4\, \leq\, 0,\, \Rightarrow\, x\, \leq\, \dfrac{5}{4}\)





is this right?
ed. ah mistake, after leaving logs I should've changed ≤ to ≥ since the base was smaller than 1

 

[tkhunny]

Hello. I am trying to tackle this inequality:

. . . . .\(\displaystyle \log_{\frac{1}{2}}(2x\, -\, 1)\, -\, \log_2(5\, -\, 3x)\, \leq\, 0\)

First, find the domain for each of the logs:

. . .\(\displaystyle \log_{\frac{1}{2}}(2x\, -\, 1):\)

. . . . .\(\displaystyle 2x\, -\, 1\, >\, 0\)
. . . . .\(\displaystyle x\, >\, \dfrac{1}{2}\)

. . .\(\displaystyle \log_2(5\, -\, 3x):\)

. . . . .\(\displaystyle 5\, -\, 3x\, >\, 0\)
. . . . .\(\displaystyle x\, <\, \dfrac{5}{3}\)

Then should I get rid of the log, by changing the base of 2nd log to 1/2?

. . . . .\(\displaystyle \log_a(b)\, =\, \dfrac{\log_c(b)}{\log_c(a)}\)

. . .\(\displaystyle \log_{\frac{1}{2}}(2x\, -\, 1)\, -\, \log_2(5\, -\, 3x)\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, \log_{\frac{1}{2}}(2x\, -\, 1)\, -\, \dfrac{\log_{\frac{1}{2}}(5\, -\, 3x)}{\log_{\frac{1}{2}}(2)}\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, \log_{\frac{1}{2}}(2x\, -\, 1)\, -\, \log_{\frac{1}{2}}(5\, -\, 3x)\, -\, \log_{\frac{1}{2}}(2)\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, 2x\, -\, 2\, -\, (5\, -\, 3x\, -\, 2)\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, 2x\, -\, 1\, -\, 5\, +\, 2x\, +\, 2\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, 5x\, -\, 4\, \leq\, 0,\, \Rightarrow\, x\, \leq\, \dfrac{5}{4}\)


View attachment 9577


is this right?

First, "get rid of the log" doesn't mean anything.

Second, why would you move from 2 to 1/2, rather than 1/2 to 2? Make your life easier, not harder.

Third, 1/2 = 2^1, so \(\displaystyle \log_{1/2}(2x-1) = -\log_{2}(2x-1)\)

Fourth, \(\displaystyle log(a)/log(b)\ne log(a)-log(b)\). That's a division of logarithms, not a logarithm of a division.

Fifth, your plan was okay. Mostly just suggestions.

 

[Dr.Peterson]

. . . . .\(\displaystyle \Rightarrow\, \log_{\frac{1}{2}}(2x\, -\, 1)\, -\, \log_{\frac{1}{2}}(5\, -\, 3x)\, -\, \log_{\frac{1}{2}}(2)\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, 2x\, -\, 2\, -\, (5\, -\, 3x\, -\, 2)\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, 2x\, -\, 1\, -\, 5\, +\, 2x\, +\, 2\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, 5x\, -\, 4\, \leq\, 0,\, \Rightarrow\, x\, \leq\, \dfrac{5}{4}\)

is this right?
ed. ah mistake, after leaving logs I should've changed ≤ to ≥ since the base was smaller than 1

An additional comment. The part above is just wrong. You can't put the arguments of logs being added or subtracted together to make a single expression, regardless of the direction of the inequality.

The domain is just the intersection of the domains of the terms; you got that right from the start, 1/2 < x < 5/3. There is no interaction such as you seem to be supposing.

Try the rest of it again.

 

[JeffM]

Hello. I am trying to tackle this inequality:

. . . . .\(\displaystyle \log_{\frac{1}{2}}(2x\, -\, 1)\, -\, \log_2(5\, -\, 3x)\, \leq\, 0\)

First, find the domain for each of the logs:

. . .\(\displaystyle \log_{\frac{1}{2}}(2x\, -\, 1):\)

. . . . .\(\displaystyle 2x\, -\, 1\, >\, 0\)
. . . . .\(\displaystyle x\, >\, \dfrac{1}{2}\)

. . .\(\displaystyle \log_2(5\, -\, 3x):\)

. . . . .\(\displaystyle 5\, -\, 3x\, >\, 0\)
. . . . .\(\displaystyle x\, <\, \dfrac{5}{3}\)

Then should I get rid of the log, by changing the base of 2nd log to 1/2?

. . . . .\(\displaystyle \log_a(b)\, =\, \dfrac{\log_c(b)}{\log_c(a)}\)

. . .\(\displaystyle \log_{\frac{1}{2}}(2x\, -\, 1)\, -\, \log_2(5\, -\, 3x)\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, \log_{\frac{1}{2}}(2x\, -\, 1)\, -\, \dfrac{\log_{\frac{1}{2}}(5\, -\, 3x)}{\log_{\frac{1}{2}}(2)}\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, \log_{\frac{1}{2}}(2x\, -\, 1)\, -\, \log_{\frac{1}{2}}(5\, -\, 3x)\, -\, \log_{\frac{1}{2}}(2)\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, 2x\, -\, 2\, -\, (5\, -\, 3x\, -\, 2)\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, 2x\, -\, 1\, -\, 5\, +\, 2x\, +\, 2\, \leq\, 0\)

. . . . .\(\displaystyle \Rightarrow\, 5x\, -\, 4\, \leq\, 0,\, \Rightarrow\, x\, \leq\, \dfrac{5}{4}\)

View attachment 9577

is this right?
ed. ah mistake, after leaving logs I should've changed ≤ to ≥ since the base was smaller than 1

I too shall make a few points. You were, with one exception, generally on the right track.

Unlike tkhunny, I would not say that getting rid of logarithms doesn't mean "anything."

\(\displaystyle log_a(b) = log_a(c) \iff b = c > 0.\)

That seems to deserve the name of "getting rid of logarithms." And that process is important in lots of logarithmic equations.

Notice that, to "get rid of logarithms" in that sense, the bases must be the same. And so your first step was, correctly, to change the base of one logarithm to equal that of the other logarithm. Quite right, but that process is called "change of base."

And obviously you know that \(\displaystyle u* log_a(w)= log_a(w^u).\)

\(\displaystyle \therefore \dfrac{1}{u} * log_a(w) = log_a(w^{(1/u)}) = log_a(\sqrt{w}).\)

That rule does not have an exception if u is a logarithm.

Finally, I find that it is frequently convenient to solve a corresponding equation before tackling an inequation. Things like worry about whether < switches to > do not arise in equations. So I might start with solving

\(\displaystyle log_{\dfrac{1}{2}}(2x-1) - log_2(5-3x) = 0.\)

Once you find where the two expressions are equal, it is easy enough to figure what happens to either side of that point.

 

[Trzewikodziobowna]

So many simple mistakes. Thank you very much for the advice.

So I can solve it like this:

 

[Steven G]

So many simple mistakes. Thank you very much for the advice.

So I can solve it like this:

View attachment 9578

I would not use the change of base theorem to solve this problem. I would rather use the following theorem. log_a(b) = log_(1/a)(1/b). In your case log_(1/2)(2x-1) = log₂(2x-1)^-1 = log₂ 1/(2x-1). This way both logs are base b and you do not have to get a fraction of logs.

Then log_(1/2)(2x-1) - log₂(5-3x) = log₂ 1/(2x-1) - log₂(5-3x) = log₂ 1/[(2x-1)(5-3x)] < 0 or log₂ (2x-1)(5-3x) > 0 or (2x-1)(5-3x)>1. Continue from here. What you did to find the domain was all correct.

 

[tkhunny]

I
Unlike tkhunny, I would not say that getting rid of logarithms doesn't mean "anything."

\(\displaystyle log_a(b) = log_a(c) \iff b = c > 0.\)

The fact that you have used properties of the logarithmic expression to conclude that b = c does not mean that you have gotten rid of the logarithms. The logarithms are still there. If you ignore that you started with the logarithms, you may change the Domain of the problem.

 

[Steven G]

The fact that you have used properties of the logarithmic expression to conclude that b = c does not mean that you have gotten rid of the logarithms. The logarithms are still there. If you ignore that you started with the logarithms, you may change the Domain of the problem.

I have to agree with that logic.

 

[tkhunny]

I have to agree with that logic.

I wish the rest of the world could see that most of the time, if we just talk a little, we discover that we really are close to being on the same page.

 

[JeffM]

The fact that you have used properties of the logarithmic expression to conclude that b = c does not mean that you have gotten rid of the logarithms. The logarithms are still there. If you ignore that you started with the logarithms, you may change the Domain of the problem.

Had I said \(\displaystyle log_a(b) = log_a(c) \iff b = c\), I would grant that you had a valid point without further ado.

That, however, is not what I said. Would you please be so kind as to give me an example where confusion of domains would arise from what I actually said, namely

\(\displaystyle log_a(b) = log_a(c) \iff b = c > 0.\)

 

[tkhunny]

Had I said \(\displaystyle log_a(b) = log_a(c) \iff b = c\), I would grant that you had a valid point without further ado.

That, however, is not what I said. Would you please be so kind as to give me an example where confusion of domains would arise from what I actually said, namely

\(\displaystyle log_a(b) = log_a(b) \iff b = c > 0.\)

Pedagogical risk, that's all. I want the student to stay careful from the beginning and never become complacent and never practice too-casual language. It's a personal emphasis. Others may disagree.

 

[JeffM]

Pedagogical risk, that's all. I want the student to stay careful from the beginning and never become complacent and never practice too-casual language. It's a personal emphasis. Others may disagree.

OK. Pedagogical risk is a concept I fully subscribe to. But it leaves open a question that I pose without meaning to be argumentative.

It's a little strange in the context of this thread, where the student was actually talking about change of base, but what, if anything, is a good descriptor for the equivalence

\(\displaystyle log_a(b) = log_a(c) \iff b = c > 0.\)

It is a consequence of log functions being monotonic functions, but that doesn't give us a name for the consequence.

\(\displaystyle b > 0 < c \text { and } log_a(b) = log_a(c) \implies b = c.\)

That consequence is one of the reasons that log functions can be so useful.

 

[tkhunny]

OK. Pedagogical risk is a concept I fully subscribe to. But it leaves open a question that I pose without meaning to be argumentative.

It's a little strange in the context of this thread, where the student was actually talking about change of base, but what, if anything, is a good descriptor for the equivalence

\(\displaystyle log_a(b) = log_a(c) \iff b = c > 0.\)

It is a consequence of log functions being monotonic functions, but that doesn't give us a name for the consequence.

\(\displaystyle b > 0 < c \text { and } log_a(b) = log_a(c) \implies b = c.\)

That consequence is one of the reasons that log functions can be so useful.

No disagreement, but it seems to me to be unlikely that the beginning student will observe the completeness and subtlety of the statement \(\displaystyle b = c > 0\), and why it is complete and truly does allow \(\displaystyle \iff\), rather than just one way. Would the student have included \(\displaystyle > 0\) if left to his/her own devices? Also, the statement \(\displaystyle b = c > 0\) is a little unusual. The combination of "=" and ">" isn't particularly common.

 

[Deleted member 4993]

Had I said \(\displaystyle log_a(b) = log_a(c) \iff b = c\), I would grant that you had a valid point without further ado.

That, however, is not what I said. Would you please be so kind as to give me an example where confusion of domains would arise from what I actually said, namely

\(\displaystyle log_a(b) = log_a(b) \iff b = c > 0.\)

Jeff - you are getting too upset about little things. Your last post needs to be fixed:


log_a(b) = log_a(c) iff b = c > 0

You would have seen that I am sure - iff ......

 

[lookagain]

In your case log_(1/2)(2x-1) = log₂(2x-1)^-1 = log₂ 1/(2x-1). . . . . . This imposes a false limitation on the domain by having this quantity in the denominator.

Then log_(1/2)(2x-1) - log₂(5-3x) = log₂ 1/(2x-1) - log₂(5-3x) = log₂ 1/[(2x-1)(5-3x)] < 0 . . . . . . See above.

or log₂ (2x-1)(5-3x) > 0 . . . . . . See consequences from the above.

or (2x-1)(5-3x)>1 . . . . . . See consequences from the above.

\(\displaystyle \dfrac{log(2x - 1)}{log(1/2)} \ - \ \dfrac{log(5 - 3x)}{log(2)} \ \le \ 0\)

\(\displaystyle \dfrac{log(2x - 1)}{log(2^{-1})} \ - \ \dfrac{log(5 - 3x)}{log(2)} \ \le \ 0\)

\(\displaystyle \dfrac{log(2x - 1)}{-1*log(2)} \ - \ \dfrac{log(5 - 3x)}{log(2)} \ \le \ 0\)

\(\displaystyle \dfrac{- \ log(2x - 1)}{log(2)} \ - \ \dfrac{log(5 - 3x)}{log(2)} \ \le \ 0\)


\(\displaystyle [-log(2)]*\bigg[\dfrac{- \ log(2x - 1)}{log(2)} \ - \ \dfrac{log(5 - 3x)}{log(2)}\bigg] \ \ge \ (0)[-log(2)]\)


\(\displaystyle log(2x - 1) \ + \ log(5 - 3x) \ \ge \ 0\)

\(\displaystyle log[(2x - 1)(5 - 3x)] \ \ge \ 0\)

\(\displaystyle (2x - 1)(5 - 3x) \ \ge \ 10^0\)

\(\displaystyle (2x - 1)(5 - 3x) \ \ge \ 1\)


And continue solving, noting the restrictions of the arguments in the original inequality.


And putting the original inequality into WolframAlpha shows how the endpoints are included. It's about the eighth block of line down.
The values in the solution are in decimal (rounded/exact) form.

https://www.wolframalpha.com/input/?i=log_0.5(2x+-+1)+-+log_2(5+-+3x)+<=+0

 

[JeffM]

Jeff - you are getting too upset about little things. Your last post needs to be fixed:


log_a(b) = log_a(c) iff b = c > 0

You would have seen that I am sure - iff ......

LOL

You are right. I shall do so at once. Thanks for pointing out my error.

I am not upset. I am curious. I have no name for what is a property of monotonic functions.

 

[JeffM]

it seems to me to be unlikely that the beginning student will observe the completeness ... of \(\displaystyle b = c > 0\)

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.

 

[Deleted member 4993]

it seems to me to be unlikely that the beginning student will observe the completeness ... of \(\displaystyle b = c > 0\)

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.

Find an inverse where the value of the function is >1.:-D

 

[Steven G]

\(\displaystyle \dfrac{log(2x - 1)}{log(1/2)} \ - \ \dfrac{log(5 - 3x)}{log(2)} \ \le \ 0\)

\(\displaystyle \dfrac{log(2x - 1)}{log(2^{-1})} \ - \ \dfrac{log(5 - 3x)}{log(2)} \ \le \ 0\)

\(\displaystyle \dfrac{log(2x - 1)}{-1*log(2)} \ - \ \dfrac{log(5 - 3x)}{log(2)} \ \le \ 0\)

\(\displaystyle \dfrac{- \ log(2x - 1)}{log(2)} \ - \ \dfrac{log(5 - 3x)}{log(2)} \ \le \ 0\)


\(\displaystyle [-log(2)]*\bigg[\dfrac{- \ log(2x - 1)}{log(2)} \ - \ \dfrac{log(5 - 3x)}{log(2)}\bigg] \ \ge \ (0)[-log(2)]\)


\(\displaystyle log(2x - 1) \ + \ log(5 - 3x) \ \ge \ 0\)

\(\displaystyle log[(2x - 1)(5 - 3x)] \ \ge \ 0\)

\(\displaystyle (2x - 1)(5 - 3x) \ \ge \ 10^0\)

\(\displaystyle (2x - 1)(5 - 3x) \ \ge \ 1\)


And continue solving, noting the restrictions of the arguments in the original inequality.


And putting the original inequality into WolframAlpha shows how the endpoints are included. It's about the eighth block of line down.
The values in the solution are in decimal (rounded/exact) form.

https://www.wolframalpha.com/input/?i=log_0.5(2x+-+1)+-+log_2(5+-+3x)+<=+0

What I did was absolutely fine. I did say in my post something about the OP using their original domain that they found but you unfairly decided not to mention that.

 

[lookagain]

What I did was absolutely fine.

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.