One Logarithmic Inequality

[Vertciel]

Hello there,

I am having trouble with the following problem and I would appreciate any help or hints. My work is shown below.

Thank you very much.

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1.. For what values of \(\displaystyle x\), do we have \(\displaystyle \log_x 5 > \log_x 10\) ?

My Work :

1. I treated \(\displaystyle \log_x 5 > \log_x 10\) as \(\displaystyle b^a > c\):

Therefore:

\(\displaystyle x^{\log_x 10} > 5\)

\(\displaystyle 10 > 5\)

\(\displaystyle 2 > 1\) <-- What does this signify?

 

[o_O]

That's not true ...

For example:

\(\displaystyle log_{10} 1000 > log_{10} 100\)

\(\displaystyle \mbox{You can't say:} \: 10^{log_{10}100} > 1000\)

You have to reverse the '>' just as you would with a negative sign or a reciprocal.

Also, if you come up with a statement like '2 > 1' then that means no matter what your x values, the inequality will hold true as you will always end up with the line '2 > 1'. If you come up with an absurdity, '2 < 1' for example for all x values, what does that tell you?

 

[pka]

I don’t know how to answer this without just giving you the answer.
Here is a suggestion. Use your calculator to evaluate \(\displaystyle \log _x (5)\,\& \,\log _x (10)\) for x=1.5, 1.1,.99,.5.

Do you see what is happening?

 

[Vertciel]

Thank you for your replies.

@pka: Unfortunately, I still don't understand. If there's no other way, then I am fine with you giving me the answer.

@: If 2 < 1 for all x values, then I would know that a particular equation has no solution because 2 < 1 for all x values is simply false.

 

[stapel]

1.. For what values of \(\displaystyle x\), do we have \(\displaystyle \log_x 5 > \log_x 10\) ?

Try the change-of-base formula:

. . . . .log[sub:2lkpjsob]x[/sub:2lkpjsob](5) = ln(5)/ln(x)

. . . . .log[sub:2lkpjsob]x[/sub:2lkpjsob](10) = ln(10)/ln(x)

Clearly, ln(x) = ln(x) for all values of x for which the natural log is defined. And ln(10) > ln(5).

Since 10 = (2)(5), then ln(10) = ln(2) + ln(5). For:

. . . . .[ln(2) + ln(5)]/ln(x) = ln(2)/ln(x) + ln(5)/ln(x) < ln(5)/ln(x) = 0 + ln(5)/ln(x)

...we must have:

. . . . .ln(2)/ln(x) < 0

You know that ln(2) is positive. So what can you say about ln(x)? Then what can you say about x?

Hope that helps a bit! :wink:

Eliz.

 

[Vertciel]

Thanks for your reply, Eliz. I think that I solved the problem, but could you please check my following steps to see if I fully understood your tips?

\(\displaystyle \log_x 5 > \log_x 10\)

\(\displaystyle \frac{\log 5}{\log x} > \frac{\log 10}{\log x}\)

\(\displaystyle \frac{\log 5}{\log x} > \frac{\log 2}{\log x} + \frac{\log 5}{\log x}\)

\(\displaystyle 0 > \frac{\log 2}{\log x}\) <-- Subtracting \(\displaystyle \frac{\log 5}{\log x}\) from both sides

Since \(\displaystyle \log 2\) is positive, then \(\displaystyle \log x\) must be negative. Also, since all values from \(\displaystyle 0 < x < 1\) produce a negative number for the function \(\displaystyle f(x) \log x\), then the x must be \(\displaystyle 0 < x < 1\).

 

[pka]

\(\displaystyle \log_x 5 > \log_x 10\)
\(\displaystyle \frac{\log 5}{\log x} > \frac{\log 10}{\log x}\)
\(\displaystyle \frac{\log 5}{\log x} > \frac{\log 2}{\log x} + \frac{\log 5}{\log x}\)
\(\displaystyle 0 > \frac{\log 2}{\log x}\) <-- Subtracting \(\displaystyle \frac{\log 5}{\log x}\) from both sides
Since \(\displaystyle \log 2\) is positive, then \(\displaystyle \log x\) must be negative. Also, since all values from \(\displaystyle 0 < x < 1\) produce a negative number for the function \(\displaystyle f(x) \log x\), then the x must be \(\displaystyle 0 < x < 1\).

Your basic mistake is dividing by a negative.
Dividing by a negative, changes the sense of the inequality.
\(\displaystyle \log _{\frac{1}{2}} \left( 5 \right) \approx - {\rm{2}}{\rm{.322}}\).

 

[Vertciel]

Thanks for your reply, pka.

Are you saying that for \(\displaystyle \frac{\log 2}{\log x} < 0\), \(\displaystyle \log x\) canNOT be negative? If you are, could you please explain?

My reasoning is that if \(\displaystyle \frac{\log 2}{\log x} < 0\), then the left side must be negative to be less than 0. Otherwise, if it is positive, than it will be greater than 0.

 

[Deleted member 4993]

Hello there,

I am having trouble with the following problem and I would appreciate any help or hints. My work is shown below.

Thank you very much.

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1.. For what values of \(\displaystyle x\), do we have \(\displaystyle \log_x 5 > \log_x 10\) ?

\(\displaystyle \log_x 5 > \log_x 10\)

\(\displaystyle \log_x 5 > \log_x (5\cdot 2)\)

\(\displaystyle \log_x 5 > \log_x 5 + \log_x 2\)

\(\displaystyle 0 > \log_x 2\)

Let

\(\displaystyle \log_x 2 = - y\)

where y > 0

then

\(\displaystyle x ^ {-y} = 2\)

\(\displaystyle x ^ {y} = \frac{1}{2}\)

since right-hand-side < 1, --> x<1 (that is x is a fraction)