Exponent questions on the SAT ask which expression is equivalent to another, or ask you to rewrite a radical as a power. A handful of rules cover everything. The trouble is that the rules look similar to each other, and the answer choices are built from the versions that mix them up.
The rule
Multiply means add exponents, divide means subtract, and a power of a power means multiply, but only when the bases are identical.
Multiplying powers of the same base adds the exponents, so x to the fourth times x to the third is x to the seventh. Dividing subtracts them, so x to the ninth divided by x to the second is x to the seventh. Raising a power to another power multiplies them, so the quantity x to the fourth raised to the third power is x to the twelfth. Those three rules account for most questions.
The word same in that first sentence is load bearing. You can only add exponents when the bases match. The expression 2 to the third times 5 to the third does not simplify by adding exponents, though it does simplify by the other direction of the power rule to the quantity 2 times 5 raised to the third power, which is 10 to the third. When bases differ, look for a way to rewrite one of them, such as turning 8 into 2 to the third.
A negative exponent means a reciprocal, not a negative number. So x to the negative 2 equals 1 over x squared, which is positive whenever x is not zero. Moving a factor across the fraction bar flips the sign of its exponent, so 3 over x to the negative 4 equals 3 times x to the fourth. And any nonzero base to the zero power is 1, which is the fastest way to collapse an ugly-looking term.
Fractional exponents encode roots. The denominator is the root and the numerator is the power, so x to the two thirds is the cube root of x squared, and the square root of x is x to the one half. Converting radicals to fractional exponents is usually the fastest way to simplify an expression that mixes roots and powers, because then the three basic rules apply directly.
When the algebra looks risky, test a number. Substitute x equals 2 into the original expression and into each choice, and keep whichever matches. Avoid 0 and 1 as test values, since almost everything agrees at those inputs. If two choices still match, run a second value such as 3 to break the tie.