For example, x3{\displaystyle x^{3}} is the same as x×x×x{\displaystyle x\times x\times x}. Plugging in a number, you would have23{\displaystyle 2^{3}}=2×2×2{\displaystyle 2\times 2\times 2}=8{\displaystyle 8} Expressions to the first degree (expressions with an exponent of 1) always simplify to the base. It is like saying “x one time. ” For example, x1=x{\displaystyle x^{1}=x}. Expressions to the zero degree (expressions with an exponent of 0) always simplify to 1. For example, x0=1{\displaystyle x^{0}=1}.
This rule does not apply to numbers that have a different base. For example, you cannot simplify 23×32{\displaystyle 2^{3}\times 3^{2}}, you simply have to solve the exponents separately and then multiply the two numbers. For example, x2×x4{\displaystyle x^{2}\times x^{4}} is the same as x2+4{\displaystyle x^{2+4}}, which is the same as x6{\displaystyle x^{6}}. Plugging in a number, you would have22×24{\displaystyle 2^{2}\times 2^{4}}=22+4{\displaystyle 2^{2+4}}=26{\displaystyle 2^{6}}=2×2×2×2×2×2{\displaystyle 2\times 2\times 2\times 2\times 2\times 2}=64{\displaystyle 64}
For example, x10x5{\displaystyle {\frac {x^{10}}{x^{5}}}} is the same as x10−5{\displaystyle x^{10-5}}, which is the same as x5{\displaystyle x^{5}}. Plugging in a number, you would have 21025{\displaystyle {\frac {2^{10}}{2^{5}}}}=210−5{\displaystyle 2^{10-5}}=25{\displaystyle 2^{5}}=2×2×2×2×2{\displaystyle 2\times 2\times 2\times 2\times 2}=32{\displaystyle 32}
For example, (x2)3{\displaystyle (x^{2})^{3}} is the same as x2×3{\displaystyle x^{2\times 3}}, which is the same as x6{\displaystyle x^{6}}. Plugging in a number, you would have(22)3{\displaystyle (2^{2})^{3}}=22×3{\displaystyle 2^{2\times 3}}=26{\displaystyle 2^{6}}=2×2×2×2×2×2{\displaystyle 2\times 2\times 2\times 2\times 2\times 2}=64{\displaystyle 64}
For example, x−4{\displaystyle x^{-4}} is the same as 1x4{\displaystyle {\frac {1}{x^{4}}}}. Plugging in a number,2−4{\displaystyle 2^{-4}}=124{\displaystyle {\frac {1}{2^{4}}}}=12×2×2×2{\displaystyle {\frac {1}{2\times 2\times 2\times 2}}}=116{\displaystyle {\frac {1}{16}}}
For example, if the problem is 4(x10÷x4)÷2(x3×x2)−3x0{\displaystyle 4(x^{10}\div x^{4})\div 2(x^{3}\times x^{2})-3x^{0}}, you would first complete the calculations inside the parentheses.
For example, x10÷x4{\displaystyle x^{10}\div x^{4}} can simplify to x10−4{\displaystyle x^{10-4}}, or x6{\displaystyle x^{6}}. x3×x2{\displaystyle x^{3}\times x^{2}} can simplify to x3+2{\displaystyle x^{3+2}}, or x5{\displaystyle x^{5}}. x0{\displaystyle x^{0}} is 1, since any number to the zero power is 1. So, the simplified problem becomes 4(x6)÷2(x5)−3(1){\displaystyle 4(x^{6})\div 2(x^{5})-3(1)}.
For example, for 4(x6)÷2(x5){\displaystyle 4(x^{6})\div 2(x^{5})}, you would first divide the coefficients:4÷2=2{\displaystyle 4\div 2=2}. Then, divide the exponents:x6÷x5{\displaystyle x^{6}\div x^{5}}=x6−5{\displaystyle x^{6-5}}=x1{\displaystyle x^{1}}=x{\displaystyle x}. Since 3(1){\displaystyle 3(1)} simplifies to 3{\displaystyle 3}, the final, simplified problem is 2x−3{\displaystyle 2x-3}.
