What sort of sorcery is this?

Since $ab = a^b$ and $\dfrac {a}{b} = a^{3b}$ , we have

$a^2 = ab \times \dfrac {a}{b} = a^b \times a^{3b} = a^{4b}.$

Therefore, $b = \dfrac{1}{2}$ .

Substituting, we obtain $\dfrac{1}{2}a = a^{1/2}$ so $a^{1/2} = 2$ and $a=4$ .

Then $b^{-a} = \left (\dfrac{1}{2} \right )^{-4} = \boxed{16}$ .

By the first equation we get:

$b = a^{b - 1} \qquad (1)$

By the second equation we get:

$\frac{1}{b} = a^{3b - 1}$ $b = a^{1 - 3b} \qquad (2)$

As (1) = (2) then:

$a^{b - 1} = a^{1 - 3b}$ $b - 1 = 1 - 3b$ $b = \frac{1}{2} \qquad (3)$

Replacing (3) in (1):

$\frac{1}{2} = a^{ - \frac{1}{2}} = 4$

Therefore:

$b^{ - a} = \left( {\frac{1}{2}} \right)^{ - 4} = 16$

$\begin{aligned} \frac{a}{b} & = a^{3b} = (a^b)^3 = (ab)^3 = a^3b^3 \quad \Rightarrow b^4 & = a^{-2} \quad \Rightarrow b = a^{-\frac{1}{2}} \\ \Rightarrow ab & = a\dot{} a^{-\frac{1}{2}} = a^{\frac{1}{2}} = a^b \quad \Rightarrow b = \frac{1}{2} \\ \Rightarrow ab & = \frac{a}{2} = a^{\frac{1}{2}} \quad \Rightarrow a^{\frac{1}{2}} = 2 \quad \Rightarrow a = 4 \\ \Rightarrow b^{-a} & = \left(\frac{1}{2} \right)^{-4} = \boxed{16} \end{aligned}$

There's an easy way using log. Taking log of both equations and simplifying by log rules gives a system of equations that you can solve by adding the two rows.

log(a) + log(b) = b log(a)

log(a) - log(b) = 3b log(a)

Adding the rows gives

2 log(a) = 4b log(a)

Because a > 1, log(a)≠0, so you can divide both sides by 2 log(a), giving

1= 2b

b= 1/2

Now it is trivial to get get $a$ by substitution.

I did it a longer way using logs.

By taking the log of the first equation we get $\log\left(ab\right)=\log\left(a^b\right)\Rightarrow \log\left(a\right)+\log\left(b\right)=b\log\left(a\right)\Rightarrow \log\left(a\right)=\frac{\log\left(b\right)}{b-1}$

Since $a>1$ the second equation proves $b\neq 1$ , meaning this division is allowed.

Taking the log of the second equation we get $log\left(\frac{a}{b}\right)=\log\left(a^{3b}\right)\Rightarrow \log\left(a\right)-\log\left(b\right)=3b\log\left(a\right)$

Substituting in $\log\left(a\right)=\frac{\log\left(b\right)}{b-1}$ we get $\frac{\log\left(b\right)}{b-1}-\log\left(b\right)=3b\frac{\log\left(b\right)}{b-1}$

$b\neq 1 \therefore \log\left(b\right)\neq0\therefore \frac{1}{b-1}-1=\frac{3b}{b-1}\Rightarrow 1-b+1=3b\Rightarrow 4b=2\Rightarrow b=\frac{1}{2}$

This now gives $\log\left(a\right)=\frac{\log\left(\frac{1}{2}\right)}{\frac{1}{2}-1}=-2\log\left(\frac{1}{2}\right)=\log\left(4\right)\Rightarrow a=4$

So $b^{-a}=\left(\frac{1}{2}\right)^{-4}=2^4=\boxed{16}$

a.b=a^b ,, take( ln) for 2 terms ln(a.b)=ln(a^b) so ln(a)+ln(b)=b.ln(a),,,,,,eq#1 a÷b=a^3.b take ln so ln(a/b)=ln(a^3.b) so. ln(a)-ln(b)=3.b.ln(a) eq #2 ,,,,,eq#1+eq#2 we get 2.ln(a)=4b.ln(a) so b=.5 sub in a.b=a^b a/2=a^.5 so. √a=2 ,,,,a=4. b^-a=.5^-4=16########

first equation let it be ab=a^b........second equation let it be a/b=a^(3b)......manipulate second equation...it becomes a^(3b).b=a..........divide equation one and the manipulateD equation two......after cancelling you will get a^(1-3b)= a^(b-1).....compare power only you get 1-3b=b-1.....then you get b = 1/2........sub b=1/2 back into equation one you get 1/2a=a^(1/2).....square both sides you get a^2-4a=0.........factorise a you get a(a-4)=0....since a>1, a=4........now just put according to the question wanted you get 16