The problem asks us to fill in the missing indices in the given expressions involving exponents. We need to use the rule that $x^a * x^b = x^{a+b}$.

1. Problem Description

The problem asks us to fill in the missing indices in the given expressions involving exponents. We need to use the rule that $x^a * x^b = x^{a+b}$.

2. Solution Steps

a) $m^5 * m^3 = m^?$

$5 + 3 = 8$

$m^5 * m^3 = m^8$

b) $n^5 * n^? = n$

Since $n = n^1$, we have $5 + ? = 1$, so $? = 1 - 5 = -4$.

$n^5 * n^{-4} = n^1 = n$

c) $w * w^3 = w^?$

Since $w = w^1$, we have $1 + 3 = 4$.

$w * w^3 = w^4$

d) $g^{-2} * g^7 = g^?$

$-2 + 7 = 5$

$g^{-2} * g^7 = g^5$

e) $p * p^{-9} * p^7 = p^?$

Since $p = p^1$, we have $1 + (-9) + 7 = 1 - 9 + 7 = -1$

$p * p^{-9} * p^7 = p^{-1}$

f) $x^{\frac{1}{2}} * x^{\frac{1}{2}} = x^?$

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

$x^{\frac{1}{2}} * x^{\frac{1}{2}} = x^1 = x$

g) $u * u^{\frac{1}{2}} = u^?$

Since $u = u^1$, we have $1 + \frac{1}{2} = \frac{2}{2} + \frac{1}{2} = \frac{3}{2}$

$u * u^{\frac{1}{2}} = u^{\frac{3}{2}}$

h) $a^{-1} * a^{\frac{3}{4}} = a^?$

$-1 + \frac{3}{4} = -\frac{4}{4} + \frac{3}{4} = -\frac{1}{4}$

$a^{-1} * a^{\frac{3}{4}} = a^{-\frac{1}{4}}$

i) $g^{-2} * g^2 = g^? = ...$

$-2 + 2 = 0$

$g^{-2} * g^2 = g^0 = 1$

j) $d^{\frac{3}{2}} * d^{\frac{1}{3}} = d^?$

$\frac{3}{2} + \frac{1}{3} = \frac{9}{6} + \frac{2}{6} = \frac{11}{6}$

$d^{\frac{3}{2}} * d^{\frac{1}{3}} = d^{\frac{11}{6}}$

3. Final Answer

a) $m^8$

b) $n^{-4}$

c) $w^4$

d) $g^5$

e) $p^{-1}$

f) $x^1$

g) $u^{\frac{3}{2}}$

h) $a^{-\frac{1}{4}}$

i) $g^0$

j) $d^{\frac{11}{6}}$