If $a^b = b^a$ and $a=2b$ then find the value of $a^2+b^2$ If $a^b = b^a$ and $a=2b$ then find the value of $a^2+b^2$
My Attempt, 
$$a^b = b^a$$
$$a^b =b^{2b}$$
$$a^b =b^b.b^b$$.
Now, what should I do further? 
 A: You didn't specify what are $a,b$ ... Integers ? positive real numbers ? ...
I will assume that $a,b$ are positive real numbers, such that $a^b=b^a$ and $a=2b$.
So we have $(2b)^b=b^{2b}$, which simplifies to $2^b=b^b$, hence $b=2$ and $a=4$.
A: $a^b = b^a$, $a$, $b$ are not $= 0$.
$b\log(a)=a\log(b)$
$b\log(2b)=2b\log(b)$
$\log(2b)=2\log(b)$
$2b=b^2$
$b=2$
$a=4$
A: Continue from your solution -
$a = (b^{2b})^{\frac 1b}$
$a = b^2$
Also a = 2b.
So we have,
$b^2 = 2b$
Assuming b is real positive number.
b = 2.
a = 4.
A: taking the lagarithm of both sides we get $$b\ln(a)=a\ln(b)$$ with $$a=2b$$ follows:
$$b\ln(2b)=2b\ln(b))$$thus we have $$\ln(2b)=2\ln(b)$$ from here we get:
$$\ln (2)+\ln(b)=2\ln(b)$$ thus $$b=2$$ and $$a=4$$ and $$a^2+b^2=4+16=20$$
A: We know that $a=2b $, so we have $$a^b = b^b \cdot b^b \Rightarrow (2b)^b = b^b \cdot b^b \Rightarrow 2^b = b^b $$. Assuming $b\in \mathbb Z $, we get, $b=2$ and thus $a=4$, so thus $\boxed {a^2+b^2=20} $. Hope it helps. 
A: If we assume that $a>0$ and $b>0$, then from $a^b=b^a$ and taking log we have $b\ln a=a\ln b$. With $a=2b$, we have
$b\ln(2b)=2b\ln b$ which implies $b(\ln(2b)-2\ln b)=0$.
Hence $b=0$ or $\ln(2b)-2\ln b=0$   or $2b=b^2$ or $b(b-2)=0$
Therefore $b=0$ or $b=2$. But $b>0$ from assumption so we choose $b=2$ so $a=4$, hence $a^2+b^2=20$
