$|a^2x+a^x+2-1|ge 1$ equation for positive a [closed]

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If $|a^2x+a^x+2-1|ge 1$ for all values of a(a>0), $ne 1$. Find the domain of x.

I tried to substitute $a^x=t$ and used the following $|
t^2+a^2t-1|ge 1$ but it is getting complicated.

inequality

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asked Jul 27 at 6:58

Samar Imam Zaidi

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closed as off-topic by Carl Mummert, Brian Borchers, Mostafa Ayaz, John B, John Ma Jul 28 at 18:52

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Brian Borchers, Mostafa Ayaz, John B, John Ma

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If $|a^2x+a^x+2-1|ge 1$ for all values of a(a>0), $ne 1$. Find the domain of x.

I tried to substitute $a^x=t$ and used the following $|
t^2+a^2t-1|ge 1$ but it is getting complicated.

inequality

share|cite|improve this question

asked Jul 27 at 6:58

Samar Imam Zaidi

1,053316

closed as off-topic by Carl Mummert, Brian Borchers, Mostafa Ayaz, John B, John Ma Jul 28 at 18:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Brian Borchers, Mostafa Ayaz, John B, John Ma

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

If $|a^2x+a^x+2-1|ge 1$ for all values of a(a>0), $ne 1$. Find the domain of x.

I tried to substitute $a^x=t$ and used the following $|
t^2+a^2t-1|ge 1$ but it is getting complicated.

inequality

share|cite|improve this question

asked Jul 27 at 6:58

Samar Imam Zaidi

1,053316

If $|a^2x+a^x+2-1|ge 1$ for all values of a(a>0), $ne 1$. Find the domain of x.

I tried to substitute $a^x=t$ and used the following $|
t^2+a^2t-1|ge 1$ but it is getting complicated.

inequality

share|cite|improve this question

asked Jul 27 at 6:58

Samar Imam Zaidi

1,053316

share|cite|improve this question

share|cite|improve this question

asked Jul 27 at 6:58

Samar Imam Zaidi

1,053316

asked Jul 27 at 6:58

Samar Imam Zaidi

1,053316

asked Jul 27 at 6:58

Samar Imam Zaidi

1,053316

1,053316

closed as off-topic by Carl Mummert, Brian Borchers, Mostafa Ayaz, John B, John Ma Jul 28 at 18:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Brian Borchers, Mostafa Ayaz, John B, John Ma

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Carl Mummert, Brian Borchers, Mostafa Ayaz, John B, John Ma Jul 28 at 18:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Brian Borchers, Mostafa Ayaz, John B, John Ma

If this question can be reworded to fit the rules in the help center, please edit the question.

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add a comment | 

2 Answers
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up vote
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Assuming that $a, t > 0$,

If $t^2 + a^2 t - 1 < 0$, then

beginalign
|t^2 + a^2 t - 1| &ge 1 \
1 - a^2 t - t^2 &ge 1 \
t^2 + a^2 t &< 0
endalign

Which has no positive real solution.

If $t^2 + a^2 t - 1 ge 0$, then

beginalign
|t^2 + a^2 t - 1| &ge 1 \
t^2 + a^2 t - 1 &ge 1 \
(t + frac 12 a^2)^2 &ge 2 + frac 14 a^4 \
t + frac 12 a^2 &ge frac 12sqrt8 + a^4 \
t &ge frac 12left(sqrt8 + a^4 - a^2 right) \
t &ge dfrac4sqrt8 + a^4 + a^2 \
a^x &ge dfrac4sqrt8 + a^4 + a^2 \
x ln a &ge ln 4 -ln(sqrt8 + a^4 + a^2) \
x &ge dfracln 4 - ln(sqrt8 + a^4 + a^2)ln a \
&textOR \
a^x &ge frac 12left(sqrt8 + a^4 - a^2 right) \
x ln a &ge lnleft(sqrt8 + a^4 - a^2 right) - ln 2 \
x &ge dfraclnleft(sqrt8 + a^4 - a^2 right)ln a
endalign

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edited Jul 27 at 20:56

answered Jul 27 at 8:17

steven gregory

16.4k22055

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The answer is $-2<x<0$. First note that $|a^2x+a^x+2-1| geq 1$ iff $a^2x+a^x+2-1 geq 1$ or $a^2x+a^x+2-1 leq -1$. But the second possibility is ruled out because $a^2x+a^x+2$ is always positive. Hence we have to find $x$ such that $a^2x+a^x+2 geq 2$. If $xgeq 0$ then we get a contradiction to the inequality by letting $a$ decrease to $0$. Hence $x<0$. If $xleq -2$ we get a contradiction by letting $a to infty$. Hence $-2<x<0$. To prove that the inequality holds for all $x$ in this range for all $a>0$ you can check that $a^2x+a^x+2$ is a decreasing function of $a$ in $(0,1)$ and an increasing function in $(1,infty )$. [For this check that both terms are decreasing when $a<1$ and both terms are increasing when $a>1$]. Hence its minimum value with respect to $a$ is attained at $a=1$. Since $a^2x+a^x+2 geq 2$ is satisfied when $a=1$ we see that it holds for all $a>0$.

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answered Jul 27 at 7:47

Kavi Rama Murthy

20k2829

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Assuming that $a, t > 0$,

If $t^2 + a^2 t - 1 < 0$, then

beginalign
|t^2 + a^2 t - 1| &ge 1 \
1 - a^2 t - t^2 &ge 1 \
t^2 + a^2 t &< 0
endalign

Which has no positive real solution.

If $t^2 + a^2 t - 1 ge 0$, then

beginalign
|t^2 + a^2 t - 1| &ge 1 \
t^2 + a^2 t - 1 &ge 1 \
(t + frac 12 a^2)^2 &ge 2 + frac 14 a^4 \
t + frac 12 a^2 &ge frac 12sqrt8 + a^4 \
t &ge frac 12left(sqrt8 + a^4 - a^2 right) \
t &ge dfrac4sqrt8 + a^4 + a^2 \
a^x &ge dfrac4sqrt8 + a^4 + a^2 \
x ln a &ge ln 4 -ln(sqrt8 + a^4 + a^2) \
x &ge dfracln 4 - ln(sqrt8 + a^4 + a^2)ln a \
&textOR \
a^x &ge frac 12left(sqrt8 + a^4 - a^2 right) \
x ln a &ge lnleft(sqrt8 + a^4 - a^2 right) - ln 2 \
x &ge dfraclnleft(sqrt8 + a^4 - a^2 right)ln a
endalign

share|cite|improve this answer

edited Jul 27 at 20:56

answered Jul 27 at 8:17

steven gregory

16.4k22055

add a comment | 

up vote
1
down vote

Assuming that $a, t > 0$,

If $t^2 + a^2 t - 1 < 0$, then

beginalign
|t^2 + a^2 t - 1| &ge 1 \
1 - a^2 t - t^2 &ge 1 \
t^2 + a^2 t &< 0
endalign

Which has no positive real solution.

If $t^2 + a^2 t - 1 ge 0$, then

beginalign
|t^2 + a^2 t - 1| &ge 1 \
t^2 + a^2 t - 1 &ge 1 \
(t + frac 12 a^2)^2 &ge 2 + frac 14 a^4 \
t + frac 12 a^2 &ge frac 12sqrt8 + a^4 \
t &ge frac 12left(sqrt8 + a^4 - a^2 right) \
t &ge dfrac4sqrt8 + a^4 + a^2 \
a^x &ge dfrac4sqrt8 + a^4 + a^2 \
x ln a &ge ln 4 -ln(sqrt8 + a^4 + a^2) \
x &ge dfracln 4 - ln(sqrt8 + a^4 + a^2)ln a \
&textOR \
a^x &ge frac 12left(sqrt8 + a^4 - a^2 right) \
x ln a &ge lnleft(sqrt8 + a^4 - a^2 right) - ln 2 \
x &ge dfraclnleft(sqrt8 + a^4 - a^2 right)ln a
endalign

share|cite|improve this answer

edited Jul 27 at 20:56

answered Jul 27 at 8:17

steven gregory

16.4k22055

add a comment | 

up vote
1
down vote

up vote
1
down vote

Assuming that $a, t > 0$,

If $t^2 + a^2 t - 1 < 0$, then

beginalign
|t^2 + a^2 t - 1| &ge 1 \
1 - a^2 t - t^2 &ge 1 \
t^2 + a^2 t &< 0
endalign

Which has no positive real solution.

If $t^2 + a^2 t - 1 ge 0$, then

beginalign
|t^2 + a^2 t - 1| &ge 1 \
t^2 + a^2 t - 1 &ge 1 \
(t + frac 12 a^2)^2 &ge 2 + frac 14 a^4 \
t + frac 12 a^2 &ge frac 12sqrt8 + a^4 \
t &ge frac 12left(sqrt8 + a^4 - a^2 right) \
t &ge dfrac4sqrt8 + a^4 + a^2 \
a^x &ge dfrac4sqrt8 + a^4 + a^2 \
x ln a &ge ln 4 -ln(sqrt8 + a^4 + a^2) \
x &ge dfracln 4 - ln(sqrt8 + a^4 + a^2)ln a \
&textOR \
a^x &ge frac 12left(sqrt8 + a^4 - a^2 right) \
x ln a &ge lnleft(sqrt8 + a^4 - a^2 right) - ln 2 \
x &ge dfraclnleft(sqrt8 + a^4 - a^2 right)ln a
endalign

share|cite|improve this answer

edited Jul 27 at 20:56

answered Jul 27 at 8:17

steven gregory

16.4k22055

Assuming that $a, t > 0$,

If $t^2 + a^2 t - 1 < 0$, then

beginalign
|t^2 + a^2 t - 1| &ge 1 \
1 - a^2 t - t^2 &ge 1 \
t^2 + a^2 t &< 0
endalign

Which has no positive real solution.

If $t^2 + a^2 t - 1 ge 0$, then

beginalign
|t^2 + a^2 t - 1| &ge 1 \
t^2 + a^2 t - 1 &ge 1 \
(t + frac 12 a^2)^2 &ge 2 + frac 14 a^4 \
t + frac 12 a^2 &ge frac 12sqrt8 + a^4 \
t &ge frac 12left(sqrt8 + a^4 - a^2 right) \
t &ge dfrac4sqrt8 + a^4 + a^2 \
a^x &ge dfrac4sqrt8 + a^4 + a^2 \
x ln a &ge ln 4 -ln(sqrt8 + a^4 + a^2) \
x &ge dfracln 4 - ln(sqrt8 + a^4 + a^2)ln a \
&textOR \
a^x &ge frac 12left(sqrt8 + a^4 - a^2 right) \
x ln a &ge lnleft(sqrt8 + a^4 - a^2 right) - ln 2 \
x &ge dfraclnleft(sqrt8 + a^4 - a^2 right)ln a
endalign

share|cite|improve this answer

edited Jul 27 at 20:56

answered Jul 27 at 8:17

steven gregory

16.4k22055

share|cite|improve this answer

share|cite|improve this answer

edited Jul 27 at 20:56

edited Jul 27 at 20:56

edited Jul 27 at 20:56

answered Jul 27 at 8:17

steven gregory

16.4k22055

answered Jul 27 at 8:17

steven gregory

16.4k22055

answered Jul 27 at 8:17

steven gregory

16.4k22055

16.4k22055

add a comment | 

add a comment | 

up vote
0
down vote

The answer is $-2<x<0$. First note that $|a^2x+a^x+2-1| geq 1$ iff $a^2x+a^x+2-1 geq 1$ or $a^2x+a^x+2-1 leq -1$. But the second possibility is ruled out because $a^2x+a^x+2$ is always positive. Hence we have to find $x$ such that $a^2x+a^x+2 geq 2$. If $xgeq 0$ then we get a contradiction to the inequality by letting $a$ decrease to $0$. Hence $x<0$. If $xleq -2$ we get a contradiction by letting $a to infty$. Hence $-2<x<0$. To prove that the inequality holds for all $x$ in this range for all $a>0$ you can check that $a^2x+a^x+2$ is a decreasing function of $a$ in $(0,1)$ and an increasing function in $(1,infty )$. [For this check that both terms are decreasing when $a<1$ and both terms are increasing when $a>1$]. Hence its minimum value with respect to $a$ is attained at $a=1$. Since $a^2x+a^x+2 geq 2$ is satisfied when $a=1$ we see that it holds for all $a>0$.

share|cite|improve this answer

answered Jul 27 at 7:47

Kavi Rama Murthy

20k2829

add a comment | 

up vote
0
down vote

The answer is $-2<x<0$. First note that $|a^2x+a^x+2-1| geq 1$ iff $a^2x+a^x+2-1 geq 1$ or $a^2x+a^x+2-1 leq -1$. But the second possibility is ruled out because $a^2x+a^x+2$ is always positive. Hence we have to find $x$ such that $a^2x+a^x+2 geq 2$. If $xgeq 0$ then we get a contradiction to the inequality by letting $a$ decrease to $0$. Hence $x<0$. If $xleq -2$ we get a contradiction by letting $a to infty$. Hence $-2<x<0$. To prove that the inequality holds for all $x$ in this range for all $a>0$ you can check that $a^2x+a^x+2$ is a decreasing function of $a$ in $(0,1)$ and an increasing function in $(1,infty )$. [For this check that both terms are decreasing when $a<1$ and both terms are increasing when $a>1$]. Hence its minimum value with respect to $a$ is attained at $a=1$. Since $a^2x+a^x+2 geq 2$ is satisfied when $a=1$ we see that it holds for all $a>0$.

share|cite|improve this answer

answered Jul 27 at 7:47

Kavi Rama Murthy

20k2829

add a comment | 

up vote
0
down vote

up vote
0
down vote

The answer is $-2<x<0$. First note that $|a^2x+a^x+2-1| geq 1$ iff $a^2x+a^x+2-1 geq 1$ or $a^2x+a^x+2-1 leq -1$. But the second possibility is ruled out because $a^2x+a^x+2$ is always positive. Hence we have to find $x$ such that $a^2x+a^x+2 geq 2$. If $xgeq 0$ then we get a contradiction to the inequality by letting $a$ decrease to $0$. Hence $x<0$. If $xleq -2$ we get a contradiction by letting $a to infty$. Hence $-2<x<0$. To prove that the inequality holds for all $x$ in this range for all $a>0$ you can check that $a^2x+a^x+2$ is a decreasing function of $a$ in $(0,1)$ and an increasing function in $(1,infty )$. [For this check that both terms are decreasing when $a<1$ and both terms are increasing when $a>1$]. Hence its minimum value with respect to $a$ is attained at $a=1$. Since $a^2x+a^x+2 geq 2$ is satisfied when $a=1$ we see that it holds for all $a>0$.

share|cite|improve this answer

answered Jul 27 at 7:47

Kavi Rama Murthy

20k2829

The answer is $-2<x<0$. First note that $|a^2x+a^x+2-1| geq 1$ iff $a^2x+a^x+2-1 geq 1$ or $a^2x+a^x+2-1 leq -1$. But the second possibility is ruled out because $a^2x+a^x+2$ is always positive. Hence we have to find $x$ such that $a^2x+a^x+2 geq 2$. If $xgeq 0$ then we get a contradiction to the inequality by letting $a$ decrease to $0$. Hence $x<0$. If $xleq -2$ we get a contradiction by letting $a to infty$. Hence $-2<x<0$. To prove that the inequality holds for all $x$ in this range for all $a>0$ you can check that $a^2x+a^x+2$ is a decreasing function of $a$ in $(0,1)$ and an increasing function in $(1,infty )$. [For this check that both terms are decreasing when $a<1$ and both terms are increasing when $a>1$]. Hence its minimum value with respect to $a$ is attained at $a=1$. Since $a^2x+a^x+2 geq 2$ is satisfied when $a=1$ we see that it holds for all $a>0$.

share|cite|improve this answer

answered Jul 27 at 7:47

Kavi Rama Murthy

20k2829

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 7:47

Kavi Rama Murthy

20k2829

answered Jul 27 at 7:47

Kavi Rama Murthy

20k2829

answered Jul 27 at 7:47

Kavi Rama Murthy

20k2829

20k2829

add a comment | 

add a comment |