How to solve the inequality with logarithm?
$begingroup$
The inequality is: $$frac{1}{log_{(x-1)}frac{x}{20}}ge-1$$
I made a plot of the function $f(x)=frac{1}{log_{(x-1)}frac{x}{20}}+1$ and it looks the answer is $xin(1,5]cup(20,+infty)$. Using logarithm definition I conclude that the argument $frac{x}{20}>0$, the base $x-1>0$ and $x-1ne1$. Also I made a couple of transformations of initial inequality to retrieve definite answer but it doesn't quite match what I see on the plot.
I am looking for a specific solution steps required to get the answer.
inequality logarithms fractions
$endgroup$
add a comment |
$begingroup$
The inequality is: $$frac{1}{log_{(x-1)}frac{x}{20}}ge-1$$
I made a plot of the function $f(x)=frac{1}{log_{(x-1)}frac{x}{20}}+1$ and it looks the answer is $xin(1,5]cup(20,+infty)$. Using logarithm definition I conclude that the argument $frac{x}{20}>0$, the base $x-1>0$ and $x-1ne1$. Also I made a couple of transformations of initial inequality to retrieve definite answer but it doesn't quite match what I see on the plot.
I am looking for a specific solution steps required to get the answer.
inequality logarithms fractions
$endgroup$
$begingroup$
Hint: Assuming that by $log_{(a)}(b)$ you denote the logarithm of $b$ in base $a$, express $log_{(a)}(b)$ in terms of the natural logarithms ($ln$) of $a$ and $b$.
$endgroup$
– Bill O'Haran
Dec 17 '18 at 20:04
$begingroup$
I made all standard transformations, what I am struggle with is specific details on how to get an answer that matches the one that got from the plot. Or if it's incorrect how to get the correct one. Can you describe all solution steps?
$endgroup$
– user627320
Dec 18 '18 at 4:23
$begingroup$
Please see how my answer fits your needs.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 9:32
add a comment |
$begingroup$
The inequality is: $$frac{1}{log_{(x-1)}frac{x}{20}}ge-1$$
I made a plot of the function $f(x)=frac{1}{log_{(x-1)}frac{x}{20}}+1$ and it looks the answer is $xin(1,5]cup(20,+infty)$. Using logarithm definition I conclude that the argument $frac{x}{20}>0$, the base $x-1>0$ and $x-1ne1$. Also I made a couple of transformations of initial inequality to retrieve definite answer but it doesn't quite match what I see on the plot.
I am looking for a specific solution steps required to get the answer.
inequality logarithms fractions
$endgroup$
The inequality is: $$frac{1}{log_{(x-1)}frac{x}{20}}ge-1$$
I made a plot of the function $f(x)=frac{1}{log_{(x-1)}frac{x}{20}}+1$ and it looks the answer is $xin(1,5]cup(20,+infty)$. Using logarithm definition I conclude that the argument $frac{x}{20}>0$, the base $x-1>0$ and $x-1ne1$. Also I made a couple of transformations of initial inequality to retrieve definite answer but it doesn't quite match what I see on the plot.
I am looking for a specific solution steps required to get the answer.
inequality logarithms fractions
inequality logarithms fractions
edited Dec 18 '18 at 4:25
asked Dec 17 '18 at 19:56
user627320
$begingroup$
Hint: Assuming that by $log_{(a)}(b)$ you denote the logarithm of $b$ in base $a$, express $log_{(a)}(b)$ in terms of the natural logarithms ($ln$) of $a$ and $b$.
$endgroup$
– Bill O'Haran
Dec 17 '18 at 20:04
$begingroup$
I made all standard transformations, what I am struggle with is specific details on how to get an answer that matches the one that got from the plot. Or if it's incorrect how to get the correct one. Can you describe all solution steps?
$endgroup$
– user627320
Dec 18 '18 at 4:23
$begingroup$
Please see how my answer fits your needs.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 9:32
add a comment |
$begingroup$
Hint: Assuming that by $log_{(a)}(b)$ you denote the logarithm of $b$ in base $a$, express $log_{(a)}(b)$ in terms of the natural logarithms ($ln$) of $a$ and $b$.
$endgroup$
– Bill O'Haran
Dec 17 '18 at 20:04
$begingroup$
I made all standard transformations, what I am struggle with is specific details on how to get an answer that matches the one that got from the plot. Or if it's incorrect how to get the correct one. Can you describe all solution steps?
$endgroup$
– user627320
Dec 18 '18 at 4:23
$begingroup$
Please see how my answer fits your needs.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 9:32
$begingroup$
Hint: Assuming that by $log_{(a)}(b)$ you denote the logarithm of $b$ in base $a$, express $log_{(a)}(b)$ in terms of the natural logarithms ($ln$) of $a$ and $b$.
$endgroup$
– Bill O'Haran
Dec 17 '18 at 20:04
$begingroup$
Hint: Assuming that by $log_{(a)}(b)$ you denote the logarithm of $b$ in base $a$, express $log_{(a)}(b)$ in terms of the natural logarithms ($ln$) of $a$ and $b$.
$endgroup$
– Bill O'Haran
Dec 17 '18 at 20:04
$begingroup$
I made all standard transformations, what I am struggle with is specific details on how to get an answer that matches the one that got from the plot. Or if it's incorrect how to get the correct one. Can you describe all solution steps?
$endgroup$
– user627320
Dec 18 '18 at 4:23
$begingroup$
I made all standard transformations, what I am struggle with is specific details on how to get an answer that matches the one that got from the plot. Or if it's incorrect how to get the correct one. Can you describe all solution steps?
$endgroup$
– user627320
Dec 18 '18 at 4:23
$begingroup$
Please see how my answer fits your needs.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 9:32
$begingroup$
Please see how my answer fits your needs.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 9:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As hinted by KM101 and myself:
$$
log_ab = frac{ln b}{ln a}
$$
Then your inequality is equivalent to:
$$
frac{ln (x-1)}{lnfrac{x}{20}} geq -1
$$
Note that we must have $xneq20$ and $x>1$ for all those quantities to exist.
Now, there are two cases to consider:
$lnfrac{x}{20} > 0$ ie. $frac{x}{20}>1$ and then:
$$
ln(x-1) geq lnfrac{20}{x}
$$
(we already set $x>1$ so $xneq 0$)
and by applying $xmapsto e^x$ (which is increasing on $mathbb{R}$) on each side:
$$
x-1geq frac{20}{x}
$$
No need to do much more here since we have $frac{x}{20}>1$ ie. $x>20$, the former inequallity is always verified ($x-1>19$ and $frac{20}{x}<1$). Thus, the inequality holds for all $x>20$.
$lnfrac{x}{20} < 0$ ie. $frac{x}{20}<1$ and then:
$$
ln(x-1) leq lnfrac{20}{x}
$$
and by applying $xmapsto e^x$ (which is increasing on $mathbb{R}$) on each side:
$$
x-1leq frac{20}{x}
$$
which yields
$$
x^2 - x -20 leq 0
$$
$-4$ and $5$ are obvious roots (if not use standard methods to find them). Then the sign of the polynomial is easy to find: it is positive or zero for $xleq-4$ and $xgeq5$ and negative or zero for $-4leq xleq5$. What matters for us given the inequality is the latter case. But from former hypotheses, we must also have $x>1$ and $x<20$. Thus, the inequality holds for $1<xleq 5$.
In the end, the two distinct cases indeed yield $xin (1,5]cup(20,infty)$.
answered Dec 18 '18 at 9:32
Bill O'HaranBill O'Haran
2,5431418
$endgroup$
add a comment |
$begingroup$
Hint: Use
$$log_a b = frac{ln b}{ln a}$$
which yields
$$frac{1}{frac{ln frac{x}{20}}{ln (x-1)}} geq -1$$
and the inequality can be solved simply from here onward.
answered Dec 17 '18 at 20:18
KM101KM101
6,0251525
$endgroup$
$begingroup$
I made all standard transformations, what I am struggle with is specific details on how to get an answer that matches the one that got from the plot. Or if it's incorrect how to get the correct one. Can you describe all solution steps?
$endgroup$
– user627320
Dec 18 '18 at 4:22
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As hinted by KM101 and myself:
$$
log_ab = frac{ln b}{ln a}
$$
Then your inequality is equivalent to:
$$
frac{ln (x-1)}{lnfrac{x}{20}} geq -1
$$
Note that we must have $xneq20$ and $x>1$ for all those quantities to exist.
Now, there are two cases to consider:
$lnfrac{x}{20} > 0$ ie. $frac{x}{20}>1$ and then:
$$
ln(x-1) geq lnfrac{20}{x}
$$
(we already set $x>1$ so $xneq 0$)
and by applying $xmapsto e^x$ (which is increasing on $mathbb{R}$) on each side:
$$
x-1geq frac{20}{x}
$$
No need to do much more here since we have $frac{x}{20}>1$ ie. $x>20$, the former inequallity is always verified ($x-1>19$ and $frac{20}{x}<1$). Thus, the inequality holds for all $x>20$.
$lnfrac{x}{20} < 0$ ie. $frac{x}{20}<1$ and then:
$$
ln(x-1) leq lnfrac{20}{x}
$$
and by applying $xmapsto e^x$ (which is increasing on $mathbb{R}$) on each side:
$$
x-1leq frac{20}{x}
$$
which yields
$$
x^2 - x -20 leq 0
$$
$-4$ and $5$ are obvious roots (if not use standard methods to find them). Then the sign of the polynomial is easy to find: it is positive or zero for $xleq-4$ and $xgeq5$ and negative or zero for $-4leq xleq5$. What matters for us given the inequality is the latter case. But from former hypotheses, we must also have $x>1$ and $x<20$. Thus, the inequality holds for $1<xleq 5$.
In the end, the two distinct cases indeed yield $xin (1,5]cup(20,infty)$.
answered Dec 18 '18 at 9:32
Bill O'HaranBill O'Haran
2,5431418
$endgroup$
add a comment |
$begingroup$
As hinted by KM101 and myself:
$$
log_ab = frac{ln b}{ln a}
$$
Then your inequality is equivalent to:
$$
frac{ln (x-1)}{lnfrac{x}{20}} geq -1
$$
Note that we must have $xneq20$ and $x>1$ for all those quantities to exist.
Now, there are two cases to consider:
$lnfrac{x}{20} > 0$ ie. $frac{x}{20}>1$ and then:
$$
ln(x-1) geq lnfrac{20}{x}
$$
(we already set $x>1$ so $xneq 0$)
and by applying $xmapsto e^x$ (which is increasing on $mathbb{R}$) on each side:
$$
x-1geq frac{20}{x}
$$
No need to do much more here since we have $frac{x}{20}>1$ ie. $x>20$, the former inequallity is always verified ($x-1>19$ and $frac{20}{x}<1$). Thus, the inequality holds for all $x>20$.
$lnfrac{x}{20} < 0$ ie. $frac{x}{20}<1$ and then:
$$
ln(x-1) leq lnfrac{20}{x}
$$
and by applying $xmapsto e^x$ (which is increasing on $mathbb{R}$) on each side:
$$
x-1leq frac{20}{x}
$$
which yields
$$
x^2 - x -20 leq 0
$$
$-4$ and $5$ are obvious roots (if not use standard methods to find them). Then the sign of the polynomial is easy to find: it is positive or zero for $xleq-4$ and $xgeq5$ and negative or zero for $-4leq xleq5$. What matters for us given the inequality is the latter case. But from former hypotheses, we must also have $x>1$ and $x<20$. Thus, the inequality holds for $1<xleq 5$.
In the end, the two distinct cases indeed yield $xin (1,5]cup(20,infty)$.
answered Dec 18 '18 at 9:32
Bill O'HaranBill O'Haran
2,5431418
$endgroup$
add a comment |
$begingroup$
As hinted by KM101 and myself:
$$
log_ab = frac{ln b}{ln a}
$$
Then your inequality is equivalent to:
$$
frac{ln (x-1)}{lnfrac{x}{20}} geq -1
$$
Note that we must have $xneq20$ and $x>1$ for all those quantities to exist.
Now, there are two cases to consider:
$lnfrac{x}{20} > 0$ ie. $frac{x}{20}>1$ and then:
$$
ln(x-1) geq lnfrac{20}{x}
$$
(we already set $x>1$ so $xneq 0$)
and by applying $xmapsto e^x$ (which is increasing on $mathbb{R}$) on each side:
$$
x-1geq frac{20}{x}
$$
No need to do much more here since we have $frac{x}{20}>1$ ie. $x>20$, the former inequallity is always verified ($x-1>19$ and $frac{20}{x}<1$). Thus, the inequality holds for all $x>20$.
$lnfrac{x}{20} < 0$ ie. $frac{x}{20}<1$ and then:
$$
ln(x-1) leq lnfrac{20}{x}
$$
and by applying $xmapsto e^x$ (which is increasing on $mathbb{R}$) on each side:
$$
x-1leq frac{20}{x}
$$
which yields
$$
x^2 - x -20 leq 0
$$
$-4$ and $5$ are obvious roots (if not use standard methods to find them). Then the sign of the polynomial is easy to find: it is positive or zero for $xleq-4$ and $xgeq5$ and negative or zero for $-4leq xleq5$. What matters for us given the inequality is the latter case. But from former hypotheses, we must also have $x>1$ and $x<20$. Thus, the inequality holds for $1<xleq 5$.
In the end, the two distinct cases indeed yield $xin (1,5]cup(20,infty)$.
answered Dec 18 '18 at 9:32
Bill O'HaranBill O'Haran
2,5431418
$endgroup$
As hinted by KM101 and myself:
$$
log_ab = frac{ln b}{ln a}
$$
Then your inequality is equivalent to:
$$
frac{ln (x-1)}{lnfrac{x}{20}} geq -1
$$
Note that we must have $xneq20$ and $x>1$ for all those quantities to exist.
Now, there are two cases to consider:
$lnfrac{x}{20} > 0$ ie. $frac{x}{20}>1$ and then:
$$
ln(x-1) geq lnfrac{20}{x}
$$
(we already set $x>1$ so $xneq 0$)
and by applying $xmapsto e^x$ (which is increasing on $mathbb{R}$) on each side:
$$
x-1geq frac{20}{x}
$$
No need to do much more here since we have $frac{x}{20}>1$ ie. $x>20$, the former inequallity is always verified ($x-1>19$ and $frac{20}{x}<1$). Thus, the inequality holds for all $x>20$.
$lnfrac{x}{20} < 0$ ie. $frac{x}{20}<1$ and then:
$$
ln(x-1) leq lnfrac{20}{x}
$$
and by applying $xmapsto e^x$ (which is increasing on $mathbb{R}$) on each side:
$$
x-1leq frac{20}{x}
$$
which yields
$$
x^2 - x -20 leq 0
$$
$-4$ and $5$ are obvious roots (if not use standard methods to find them). Then the sign of the polynomial is easy to find: it is positive or zero for $xleq-4$ and $xgeq5$ and negative or zero for $-4leq xleq5$. What matters for us given the inequality is the latter case. But from former hypotheses, we must also have $x>1$ and $x<20$. Thus, the inequality holds for $1<xleq 5$.
In the end, the two distinct cases indeed yield $xin (1,5]cup(20,infty)$.
answered Dec 18 '18 at 9:32
Bill O'HaranBill O'Haran
2,5431418
answered Dec 18 '18 at 9:32
Bill O'HaranBill O'Haran
2,5431418
answered Dec 18 '18 at 9:32
Bill O'HaranBill O'Haran
2,5431418
answered Dec 18 '18 at 9:32
Bill O'HaranBill O'Haran
2,5431418
2,5431418
add a comment |
add a comment |
$begingroup$
Hint: Use
$$log_a b = frac{ln b}{ln a}$$
which yields
$$frac{1}{frac{ln frac{x}{20}}{ln (x-1)}} geq -1$$
and the inequality can be solved simply from here onward.
answered Dec 17 '18 at 20:18
KM101KM101
6,0251525
$endgroup$
$begingroup$
I made all standard transformations, what I am struggle with is specific details on how to get an answer that matches the one that got from the plot. Or if it's incorrect how to get the correct one. Can you describe all solution steps?
$endgroup$
– user627320
Dec 18 '18 at 4:22
add a comment |
$begingroup$
Hint: Use
$$log_a b = frac{ln b}{ln a}$$
which yields
$$frac{1}{frac{ln frac{x}{20}}{ln (x-1)}} geq -1$$
and the inequality can be solved simply from here onward.
answered Dec 17 '18 at 20:18
KM101KM101
6,0251525
$endgroup$
$begingroup$
I made all standard transformations, what I am struggle with is specific details on how to get an answer that matches the one that got from the plot. Or if it's incorrect how to get the correct one. Can you describe all solution steps?
$endgroup$
– user627320
Dec 18 '18 at 4:22
add a comment |
$begingroup$
Hint: Use
$$log_a b = frac{ln b}{ln a}$$
which yields
$$frac{1}{frac{ln frac{x}{20}}{ln (x-1)}} geq -1$$
and the inequality can be solved simply from here onward.
answered Dec 17 '18 at 20:18
KM101KM101
6,0251525
$endgroup$
Hint: Use
$$log_a b = frac{ln b}{ln a}$$
which yields
$$frac{1}{frac{ln frac{x}{20}}{ln (x-1)}} geq -1$$
and the inequality can be solved simply from here onward.
answered Dec 17 '18 at 20:18
KM101KM101
6,0251525
answered Dec 17 '18 at 20:18
KM101KM101
6,0251525
answered Dec 17 '18 at 20:18
KM101KM101
6,0251525
answered Dec 17 '18 at 20:18
KM101KM101
6,0251525
6,0251525
$begingroup$
I made all standard transformations, what I am struggle with is specific details on how to get an answer that matches the one that got from the plot. Or if it's incorrect how to get the correct one. Can you describe all solution steps?
$endgroup$
– user627320
Dec 18 '18 at 4:22
add a comment |
$begingroup$
I made all standard transformations, what I am struggle with is specific details on how to get an answer that matches the one that got from the plot. Or if it's incorrect how to get the correct one. Can you describe all solution steps?
$endgroup$
– user627320
Dec 18 '18 at 4:22
$begingroup$
I made all standard transformations, what I am struggle with is specific details on how to get an answer that matches the one that got from the plot. Or if it's incorrect how to get the correct one. Can you describe all solution steps?
$endgroup$
– user627320
Dec 18 '18 at 4:22
$begingroup$
I made all standard transformations, what I am struggle with is specific details on how to get an answer that matches the one that got from the plot. Or if it's incorrect how to get the correct one. Can you describe all solution steps?
$endgroup$
– user627320
Dec 18 '18 at 4:22
add a comment |
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$begingroup$
Hint: Assuming that by $log_{(a)}(b)$ you denote the logarithm of $b$ in base $a$, express $log_{(a)}(b)$ in terms of the natural logarithms ($ln$) of $a$ and $b$.
$endgroup$
– Bill O'Haran
Dec 17 '18 at 20:04
$begingroup$
I made all standard transformations, what I am struggle with is specific details on how to get an answer that matches the one that got from the plot. Or if it's incorrect how to get the correct one. Can you describe all solution steps?
$endgroup$
– user627320
Dec 18 '18 at 4:23
$begingroup$
Please see how my answer fits your needs.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 9:32