determine the value of a

**Prove It** · August 24th 2010, 01:50 AM

$ab = a^b$ and $\frac{a}{b} = a^{3b}$ .

Rearranging the second equation gives $b = \frac{a}{a^{3b}} = a^{1 - 3b}$ .

Substituting into the first gives

$a(a^{1 - 3b}) = a^b$

$a^{2 - 3b} = a^b$

$2 - 3b = b$

$2 = 4b$

$b = \frac{1}{2}$ .

Back substituting gives

$b = a^{1 - 3b}$

$\frac{1}{2} = a^{1 - \frac{3}{2}}$

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

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

$a = 4$ .

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