please help me solve this

[manngaurav]

Recently i encountered an inequality of the following type , having applied all the little knowledge that i have , and trying to solve it using various online sites i was unsuccessful , please help me understand as to how to this for x

2^x + 2^|x| >(or equal to) 2^3/2

I made two case as there is a modulus function , so one will be when x>0 and the other x<0 , I also know that the minimum value of a number and its reciprocal is 2
CASE 1( x>0)
by the definition of modulus function |x| = x
hence, 2^x + 2^x >(or equal to) 2^3/2

2.2^x >(or equal to) 2.2^1/2

2^x >(or equal to)2^1/2

this gives : x > (or equal to ) 1/2 { if i apply log still the answer is same so
i am confident that i am doing right till here}


CASE 2 (x<0)
by the definition of modulus function |x| = -x
hence,
2^x + 2^-x >(or equal to) 2^3/2
writing here is difficult so i am directly writing how far i got

log₂(2^2x + 1) - x >(or equal to) -1/2

I think i am solving this wrong as there seems to be no way to solve this equation for x , please help me with case 2 , please provide the steps so i can understand , thank you very much



NOTE: PLEASE DO NOT USE CALCULATORS AS THEY WOULD NOT PROVIDE THE STEPS AND I NEED TO UNDERSTAND IT

 

[tkhunny]

If x >= 0, your problem is \(\displaystyle 2^{x} + 2^{x} = 2^{x+1} \ge 2^{3/2}\) Is there a solution for non-negative x?

If x < 0, your problem is \(\displaystyle 2^{x} + 2^{-x} = 2^{x} + \dfrac{1}{2^{x}} = \dfrac{2^{2x} + 1}{2^{x}} \ge 2^{3/2} \implies 2^{2x} + 1 \ge 2^{x + 3/2}\). Is there a solution for negative x?

 

[manngaurav]

If x >= 0, your problem is \(\displaystyle 2^{x} + 2^{x} = 2^{x+1} \ge 2^{3/2}\) Is there a solution for non-negative x?

If x < 0, your problem is \(\displaystyle 2^{x} + 2^{-x} = 2^{x} + \dfrac{1}{2^{x}} = \dfrac{2^{2x} + 1}{2^{x}} \ge 2^{3/2} \implies 2^{2x} + 1 \ge 2^{x + 3/2}\). Is there a solution for negative x?

hello, sir

in the first case there all the x >(equal to)1/2 will satisfy the equation,
for case 2 solving your inequality i get
log₂(2^2x + 1) >(equal to) (x+3 )/2 , now what...?

 

[Deleted member 4993]

hello, sir

in the first case there all the x >(equal to)1/2 will satisfy the equation,
for case 2 solving your inequality i get
log₂(2^2x + 1) >(equal to) (x+3 )/2 , now what...?

I would follow this path:

y = 2^x and C = 2√2

then

(y²+1)/y ≥ C


y² - Cy +1 ≥ 0

y_1,2 = [C ± √(C²-4)]/2 = √2 ± 1

since x<0

√2 - 1 ≤ 2^x < 1

or

log₂(√2 - 1) ≤ x ≤ 0

Using your graphing calculator (or wolframalpha.com), you should check viability of your answers.

 

[manngaurav]

I would follow this path:

y = 2^x and C = 2√2

then

(y²+1)/y ≥ C


y² - Cy +1 ≥ 0

y_1,2 = [C ± √(C²-4)]/2 = √2 ± 1

since x<0

√2 - 1 ≤ 2^x < 1

or

log₂(√2 - 1) ≤ x ≤ 0

Using your graphing calculator (or wolframalpha.com), you should check viability of your answers.

thanks , so from case 1 and case 2 the solution should be (-infinity,log₂(√2 - 1) ) U (1/2, infinity)