Solving the Equation $2(2^a-1)a^2+2=2^{a+1}-(2^{a^2}-2)a$

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In summary, the equation $2(2^a-1)a^2+2=2^{a+1}-(2^{a^2}-2)a$ has three solutions for real numbers $a$, which are $a=0$, $a=1$, and $a=-1$. Additionally, if $x<0$, then the left side of the equation is less than $0$, while the right side is greater than $0$. Hence, the second interval for $x$ should be $-1<x<0$.
  • #1
anemone
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Solve the equation $2(2^a-1)a^2+2=2^{a+1}-(2^{a^2}-2)a$ for real numbers $a$.
 
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  • #2
anemone said:
Solve the equation $2(2^a-1)a^2+2=2^{a+1}-(2^{a^2}-2)a$ for real numbers $a$.
it is clear that $a=0$ is a solution
let :$a^2=b, 2^a=c,$ we have :
$2bc-2b+2=2c-2^ba+2a---(*)$
it is also clear that $b=1$ is a solution of (*)
this gives $a=\pm 1$
 
  • #3
Albert said:
it is clear that $a=0$ is a solution
let :$a^2=b, 2^a=c,$ we have :
$2bc-2b+2=2c-2^ba+2a---(*)$
it is also clear that $b=1$ is a solution of (*)
this gives $a=\pm 1$

Thanks for participating, Albert! :D

But, can you show us more explicitly how those three are the only solutions to the problem? :eek:
 
  • #4
Albert said:
it is clear that $a=0$ is a solution
let :$a^2=b, 2^a=c,$ we have :
$2bc-2b+2=2c-2^ba+2a---(*)$
it is also clear that $b=1$ is a solution of (*)
this gives $a=\pm 1$
continue:
it is clear that $a=0$ is a solution
let :$a^2=b, 2^a=c,$ we have :
$2bc-2b+2=2c-2^ba+2a---(*)$
it is also clear that $b=1$ is a solution of (*)
this gives $a=\pm 1$
now let $b=1+x>0$
and (*) becomes:
$x(2^a-1)=-a(2^x-1)$
or:
$x(2^\sqrt{1+x}-1)=-\sqrt{1+x}(2^x-1)----(**)$
(with 1+x>0)
if $x>0$ then the left side of (**)>0 ,but the right side of (**)<0---(1) and:
if $x<0$ then the left side of (**)<0,but the right side of (**)>0---(2)
from (1)(2):
$x=0,b=1, and, \,\, a=\pm 1$
$\therefore a=0,\pm 1$ those three are the only solutions to this problem
 
  • #5
Thanks for your continued effort for submitting a complete solution, Albert! :)

I guess the second interval of $x$ as you stated in
if $x<0$ then the left side of (**)<0,but the right side of (**)>0

it should be $-1<x<0$ instead. :D

I will post the other solution that I want to share with MHB later.
 
  • #6
anemone said:
Thanks for your continued effort for submitting a complete solution, Albert! :)

I guess the second interval of $x$ as you stated in

it should be $-1<x<0$ instead. :D

I will post the other solution that I want to share with MHB later.
if x<0 ,for x+1>0, it is clear that x>-1
 

FAQ: Solving the Equation $2(2^a-1)a^2+2=2^{a+1}-(2^{a^2}-2)a$

What is the purpose of solving this equation?

The purpose of solving this equation is to find the value of variable 'a' that satisfies the equation. This can help us understand the relationship between the different terms in the equation and can be useful in solving similar equations in the future.

How do I approach solving this equation?

One approach to solving this equation is to simplify the terms on both sides of the equation and then use algebraic techniques such as factoring, expanding, and rearranging to isolate the variable 'a'. Another approach is to use a calculator or computer program to graph the equation and visually determine the solution.

What are the potential solutions to this equation?

The potential solutions to this equation can be any real number, as long as it satisfies the equation. However, some values of 'a' may result in an undefined or imaginary solution. It is important to check for extraneous solutions and restrictions on the domain of the equation.

Can this equation be solved without using algebra?

Yes, this equation can be solved without using algebra by graphing the equation and finding the intersection point between the two sides. This method can also be done using a calculator or computer program. However, using algebraic techniques can provide a more precise solution.

How can solving this equation be applied in real life?

Solving this equation can be applied in real life situations where there are multiple variables and relationships between them. For example, it can be used in finance to calculate compound interest or in physics to calculate the trajectory of a projectile. It can also be used in engineering to optimize designs or in statistics to model data.

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