solve the inequality for $x$ [closed]
solve the inequality for $x$
$$4^x < 2^{x+1} + 3$$
I got
$(2^x)[(2^x) - 2] < 3$
What I concluded from here, is that x should be less than 1.
How to find the final answer?
calculus
edited Dec 9 at 14:55
amWhy
191k28224439
asked Dec 9 at 14:48
Mathsaddict
1778
closed as off-topic by Nosrati, amWhy, Paul Frost, RRL, DRF Dec 9 at 20:31
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, amWhy, Paul Frost, RRL, DRF
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
solve the inequality for $x$
$$4^x < 2^{x+1} + 3$$
I got
$(2^x)[(2^x) - 2] < 3$
What I concluded from here, is that x should be less than 1.
How to find the final answer?
calculus
edited Dec 9 at 14:55
amWhy
191k28224439
asked Dec 9 at 14:48
Mathsaddict
1778
closed as off-topic by Nosrati, amWhy, Paul Frost, RRL, DRF Dec 9 at 20:31
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, amWhy, Paul Frost, RRL, DRF
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
solve the inequality for $x$
$$4^x < 2^{x+1} + 3$$
I got
$(2^x)[(2^x) - 2] < 3$
What I concluded from here, is that x should be less than 1.
How to find the final answer?
calculus
edited Dec 9 at 14:55
amWhy
191k28224439
asked Dec 9 at 14:48
Mathsaddict
1778
solve the inequality for $x$
$$4^x < 2^{x+1} + 3$$
I got
$(2^x)[(2^x) - 2] < 3$
What I concluded from here, is that x should be less than 1.
How to find the final answer?
calculus
calculus
edited Dec 9 at 14:55
amWhy
191k28224439
asked Dec 9 at 14:48
Mathsaddict
1778
edited Dec 9 at 14:55
amWhy
191k28224439
asked Dec 9 at 14:48
Mathsaddict
1778
edited Dec 9 at 14:55
amWhy
191k28224439
edited Dec 9 at 14:55
amWhy
191k28224439
edited Dec 9 at 14:55
amWhy
191k28224439
191k28224439
asked Dec 9 at 14:48
Mathsaddict
1778
asked Dec 9 at 14:48
Mathsaddict
1778
asked Dec 9 at 14:48
Mathsaddict
1778
1778
closed as off-topic by Nosrati, amWhy, Paul Frost, RRL, DRF Dec 9 at 20:31
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, amWhy, Paul Frost, RRL, DRF
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Nosrati, amWhy, Paul Frost, RRL, DRF Dec 9 at 20:31
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, amWhy, Paul Frost, RRL, DRF
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
Hint: Manipulating the inequality a bit yields
$$4^x < 2^{x+1}+3$$
$$4^x-2^{x+1}-3 < 0$$
$$2^{2x}-2^{x+1}-3 < 0$$
$$2^{2x}-2(2^x)-3 < 0$$
from which you can make the substitution $t = 2^x$.
$$t^2-2t-3 < 0$$
Can you continue from here?
answered Dec 9 at 14:57
KM101
4,836421
Yes, I solved it. Thank you!
– Mathsaddict
Dec 9 at 15:02
add a comment |
$$0>(2^x)^2-2(2^x)-3=(2^x-3)(2^x+1)$$
As for real $x,2^x+1>1>0,$
$implies2^x-3<0iff2^x<3iff xleln_23$
answered Dec 9 at 14:50
lab bhattacharjee
223k15156274
add a comment |
Hint: Move everything onto one side and note that $4^x=2^{2x}$. Make a particular substitution and the problem simplifies greatly.
answered Dec 9 at 14:52
Clayton
18.9k33085
add a comment |
By completing the square,
$$4^x-2^{x+1}-3=(2^x-1)^2-4.$$
This is negative when
$$-2<2^x-1<2,$$
$$-1<2^x<3,$$
or
$$-infty<x<log_23.$$
answered Dec 9 at 15:00
Yves Daoust
124k671221
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint: Manipulating the inequality a bit yields
$$4^x < 2^{x+1}+3$$
$$4^x-2^{x+1}-3 < 0$$
$$2^{2x}-2^{x+1}-3 < 0$$
$$2^{2x}-2(2^x)-3 < 0$$
from which you can make the substitution $t = 2^x$.
$$t^2-2t-3 < 0$$
Can you continue from here?
answered Dec 9 at 14:57
KM101
4,836421
Yes, I solved it. Thank you!
– Mathsaddict
Dec 9 at 15:02
add a comment |
Hint: Manipulating the inequality a bit yields
$$4^x < 2^{x+1}+3$$
$$4^x-2^{x+1}-3 < 0$$
$$2^{2x}-2^{x+1}-3 < 0$$
$$2^{2x}-2(2^x)-3 < 0$$
from which you can make the substitution $t = 2^x$.
$$t^2-2t-3 < 0$$
Can you continue from here?
answered Dec 9 at 14:57
KM101
4,836421
Yes, I solved it. Thank you!
– Mathsaddict
Dec 9 at 15:02
add a comment |
Hint: Manipulating the inequality a bit yields
$$4^x < 2^{x+1}+3$$
$$4^x-2^{x+1}-3 < 0$$
$$2^{2x}-2^{x+1}-3 < 0$$
$$2^{2x}-2(2^x)-3 < 0$$
from which you can make the substitution $t = 2^x$.
$$t^2-2t-3 < 0$$
Can you continue from here?
answered Dec 9 at 14:57
KM101
4,836421
Hint: Manipulating the inequality a bit yields
$$4^x < 2^{x+1}+3$$
$$4^x-2^{x+1}-3 < 0$$
$$2^{2x}-2^{x+1}-3 < 0$$
$$2^{2x}-2(2^x)-3 < 0$$
from which you can make the substitution $t = 2^x$.
$$t^2-2t-3 < 0$$
Can you continue from here?
answered Dec 9 at 14:57
KM101
4,836421
answered Dec 9 at 14:57
KM101
4,836421
answered Dec 9 at 14:57
KM101
4,836421
answered Dec 9 at 14:57
KM101
4,836421
4,836421
Yes, I solved it. Thank you!
– Mathsaddict
Dec 9 at 15:02
add a comment |
Yes, I solved it. Thank you!
– Mathsaddict
Dec 9 at 15:02
Yes, I solved it. Thank you!
– Mathsaddict
Dec 9 at 15:02
Yes, I solved it. Thank you!
– Mathsaddict
Dec 9 at 15:02
add a comment |
$$0>(2^x)^2-2(2^x)-3=(2^x-3)(2^x+1)$$
As for real $x,2^x+1>1>0,$
$implies2^x-3<0iff2^x<3iff xleln_23$
answered Dec 9 at 14:50
lab bhattacharjee
223k15156274
add a comment |
$$0>(2^x)^2-2(2^x)-3=(2^x-3)(2^x+1)$$
As for real $x,2^x+1>1>0,$
$implies2^x-3<0iff2^x<3iff xleln_23$
answered Dec 9 at 14:50
lab bhattacharjee
223k15156274
add a comment |
$$0>(2^x)^2-2(2^x)-3=(2^x-3)(2^x+1)$$
As for real $x,2^x+1>1>0,$
$implies2^x-3<0iff2^x<3iff xleln_23$
answered Dec 9 at 14:50
lab bhattacharjee
223k15156274
$$0>(2^x)^2-2(2^x)-3=(2^x-3)(2^x+1)$$
As for real $x,2^x+1>1>0,$
$implies2^x-3<0iff2^x<3iff xleln_23$
answered Dec 9 at 14:50
lab bhattacharjee
223k15156274
answered Dec 9 at 14:50
lab bhattacharjee
223k15156274
answered Dec 9 at 14:50
lab bhattacharjee
223k15156274
answered Dec 9 at 14:50
lab bhattacharjee
223k15156274
223k15156274
add a comment |
add a comment |
Hint: Move everything onto one side and note that $4^x=2^{2x}$. Make a particular substitution and the problem simplifies greatly.
answered Dec 9 at 14:52
Clayton
18.9k33085
add a comment |
Hint: Move everything onto one side and note that $4^x=2^{2x}$. Make a particular substitution and the problem simplifies greatly.
answered Dec 9 at 14:52
Clayton
18.9k33085
add a comment |
Hint: Move everything onto one side and note that $4^x=2^{2x}$. Make a particular substitution and the problem simplifies greatly.
answered Dec 9 at 14:52
Clayton
18.9k33085
Hint: Move everything onto one side and note that $4^x=2^{2x}$. Make a particular substitution and the problem simplifies greatly.
answered Dec 9 at 14:52
Clayton
18.9k33085
answered Dec 9 at 14:52
Clayton
18.9k33085
answered Dec 9 at 14:52
Clayton
18.9k33085
answered Dec 9 at 14:52
Clayton
18.9k33085
18.9k33085
add a comment |
add a comment |
By completing the square,
$$4^x-2^{x+1}-3=(2^x-1)^2-4.$$
This is negative when
$$-2<2^x-1<2,$$
$$-1<2^x<3,$$
or
$$-infty<x<log_23.$$
answered Dec 9 at 15:00
Yves Daoust
124k671221
add a comment |
By completing the square,
$$4^x-2^{x+1}-3=(2^x-1)^2-4.$$
This is negative when
$$-2<2^x-1<2,$$
$$-1<2^x<3,$$
or
$$-infty<x<log_23.$$
answered Dec 9 at 15:00
Yves Daoust
124k671221
add a comment |
By completing the square,
$$4^x-2^{x+1}-3=(2^x-1)^2-4.$$
This is negative when
$$-2<2^x-1<2,$$
$$-1<2^x<3,$$
or
$$-infty<x<log_23.$$
answered Dec 9 at 15:00
Yves Daoust
124k671221
By completing the square,
$$4^x-2^{x+1}-3=(2^x-1)^2-4.$$
This is negative when
$$-2<2^x-1<2,$$
$$-1<2^x<3,$$
or
$$-infty<x<log_23.$$
answered Dec 9 at 15:00
Yves Daoust
124k671221
answered Dec 9 at 15:00
Yves Daoust
124k671221
answered Dec 9 at 15:00
Yves Daoust
124k671221
answered Dec 9 at 15:00
Yves Daoust
124k671221
124k671221
add a comment |
add a comment |
