Solving a System of Equations for Real Values of $a$ and $b$

[anemone]

Solve the following system for real values of $a$ and $b$:

$2^{a^2+b}+2^{a+b^2}=8$

$\sqrt{a}+\sqrt{b}=2$

 

[mente oscura]

Solve the following system for real values of $a$ and $b$:

$2^{a^2+b}+2^{a+b^2}=8$

$\sqrt{a}+\sqrt{b}=2$

Hello.

Spoiler

I do not know. At a glance:

$$a=1 \ and \ b=1$$

Regards.

 

[kaliprasad]

Hello.

Spoiler

I do not know. At a glance:

$$a=1 \ and \ b=1$$

Regards.

Spoiler

by interchanging a and b we get same set of equations so a= b is a solution set(that does not mean that other solution does not exist). taking a= b we get a=b = 1 is a solution set

 

[anemone]

Solve the following system for real values of $a$ and $b$:

$2^{a^2+b}+2^{a+b^2}=8$

$\sqrt{a}+\sqrt{b}=2$

Suggested solution by J. Chui:

Spoiler

WLOG, let $a \ge b$ so that $\sqrt{a} \ge 1 \ge \sqrt{b}$.

Suppose that $\sqrt{a}=1+u$ and $\sqrt{b}=1-u$. Then $a+b=2+2u^2 \ge 2$ and $ab=(1-u^2)^2 \le 1$. Thus, by the AM-GM inequality, we have

$8=2^{a^2+b}+2^{a+b^2} \ge 2\sqrt{2^{a^2+b+a+b^2}}\ge 2 \sqrt{2^{(a+b)(a+b+1)-2ab}} \ge 2 \sqrt{2^{2\cdot3-2\cdot1}} \ge 2^3 \ge 8$

with equality iff $a=b$.

Since equality must hold throughtout, $a=b$ and thus the only solution to the system is $(a,b)=(1,1)$.

 

[CellCoree]

To solve this system of equations, we can use substitution. From the second equation, we can solve for $\sqrt{a}$ and substitute it into the first equation.

$\sqrt{a}=2-\sqrt{b}$

Plugging this into the first equation, we get:

$2^{(2-\sqrt{b})^2+b}+2^{(2-\sqrt{b})+b^2}=8$

Expanding the exponents and simplifying, we get:

$2^{4-4\sqrt{b}+b+b}+2^{2+2b-b^2+\sqrt{b}}=8$

Simplifying further, we get:

$2^{4+2b}+2^{2+b+\sqrt{b}}=8$

Using the properties of exponents, we can rewrite this as:

$2^{2(2+b)}+2^{1+b+\sqrt{b}}=8$

Now, we can see that the bases of the exponents are the same, so we can set the exponents equal to each other.

$2+b=1+b+\sqrt{b}$

Solving for $\sqrt{b}$, we get:

$\sqrt{b}=1$

Plugging this back into our original equation, we can solve for $a$.

$\sqrt{a}=2-\sqrt{b}$

$\sqrt{a}=2-1$

$\sqrt{a}=1$

Therefore, our solutions for $a$ and $b$ are $a=1$ and $b=1$. We can verify this by plugging these values into our original equations:

$2^{1^2+1}+2^{1+1^2}=8$

$2^{2}+2^{2}=8$

$4+4=8$

$8=8$, which is a true statement.

$\sqrt{1}+\sqrt{1}=2$

$1+1=2$ which is also a true statement.

Therefore, our solution for the system of equations is $a=1$ and $b=1$.