Solve 2^{|x+2|}-|2^{x+1}-1|=2^{x+1}+1

  • Thread starter utkarshakash
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In summary, the conversation is discussing how to solve the equation 2^{|x+2|}-|2^{x+1}-1|=2^{x+1}+1. The speaker suggests using cases to handle the absolute values and converting the equation into quadratic form by assuming 2^x=t. They also mention using the fact that 2^{x+a}= 2^a 2^x.
  • #1
utkarshakash
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Homework Statement


[itex]2^{|x+2|}-|2^{x+1}-1|=2^{x+1}+1[/itex]

Homework Equations



The Attempt at a Solution


I know this is not a direct equation in quadratic but somehow I have to convert it in that form by assuming something to be another variable. I am supposing [itex]2^x=t[/itex]. But that doesn't help me as I cannot eliminate [itex]2^{|x+2|}[/itex]
 
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  • #2
The first thing I would do it set up "cases" to handle the absolute values. x+ 2 will be positive for x> -2 and [itex]2^{x+ 1}- 1> 0[/itex] for x> -1. So if x< -2, both x+ 2 and [itex]2^{x+1}-1[/itex] are negative. If -2< x< -1, x+ 2 is positive but [itex]2^{x+1}- 1[/itex] is still negative. If x> -1, both x+ 2 and [itex]2^{x+1}- 1[/itex] are positive.

Also use the fact that [itex]2^{x+ a}= 2^a 2^x[/itex].
 

Related to Solve 2^{|x+2|}-|2^{x+1}-1|=2^{x+1}+1

1. What is the equation trying to solve?

The equation is trying to solve for the value of x that satisfies the equation: 2^{|x+2|}-|2^{x+1}-1|=2^{x+1}+1.

2. How do you approach solving this equation?

One approach is to first simplify the equation by using the properties of exponents and absolute values. Then, we can try to isolate the variable x on one side of the equation.

3. Can this equation be solved algebraically?

Yes, this equation can be solved algebraically by using the properties of exponents and absolute values, as well as basic algebraic manipulation.

4. Are there any restrictions on the values of x in this equation?

Yes, there are restrictions on the values of x in this equation. Since we cannot take the logarithm of a negative number, the expressions inside the absolute value signs must be greater than or equal to 0. This means that x+2 must be greater than or equal to 0 and x+1 must be greater than or equal to 0. Therefore, the restrictions on x are: x≥-2 and x≥-1.

5. Can this equation have multiple solutions?

Yes, it is possible for this equation to have multiple solutions. We can check our solution by substituting it back into the original equation to see if it satisfies the equation. If there are multiple solutions, they will typically be expressed in terms of logarithms or using a calculator.

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