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1. [2025汕头期末]计算$a^3 · (-a)$的结果是（
A.$a^2$
B.$-a^2$
C.$a^4$
D.$-a^4$
D
）A.$a^2$
B.$-a^2$
C.$a^4$
D.$-a^4$
答案
D $a^3 · (-a) = -a^{3+1} = -a^4$.
2. 计算$10^5 × (-10)^4 × 10^6$的结果是（
A.$-10^{15}$
B.$10^{15}$
C.$10^7$
D.$-10^7$
B
）A.$-10^{15}$
B.$10^{15}$
C.$10^7$
D.$-10^7$
答案
B $10^5 × (-10)^4 × 10^6$(先把底数化统一) $= 10^5 × 10^4 × 10^6 = 10^{15}$.
3. ▶教材P23练习T1变式 [2026深圳期末] 下列计算正确的是（
A.$b^2 · b^3 = b^6$
B.$x^3 · x^3 = 2x^3$
C.$a^4 + a^2 = a^6$
D.$m · m^5 = m^6$
D
）A.$b^2 · b^3 = b^6$
B.$x^3 · x^3 = 2x^3$
C.$a^4 + a^2 = a^6$
D.$m · m^5 = m^6$
答案
D 选项A,$b^2 · b^3 = b^{2+3} = b^5$;选项B,$x^3 · x^3 = x^{3+3} = x^6$;选项C,$a^4$与$a^2$不是同类项,不能合并;选项D,$m · m^5 = m^{1+5} = m^6$.
4. 教材P23练习T2变式 计算：
(1) $10^3 · 10^{m-1} · 10^{2m+2}$；
(2) $(a-b)^3 · (b-a) · (a-b)^5$。
(1) $10^3 · 10^{m-1} · 10^{2m+2}$；
(2) $(a-b)^3 · (b-a) · (a-b)^5$。
答案
解:(1)$10^3 · 10^{m-1} · 10^{2m+2} = 10^{3+(m-1)+(2m+2)} = 10^{3m+4}$.
(2)$(a-b)^3 · (b-a) · (a-b)^5 = -(a-b)^3 · (a-b) · (a-b)^5 = -(a-b)^{3+1+5} = -(a-b)^9$.
(2)$(a-b)^3 · (b-a) · (a-b)^5 = -(a-b)^3 · (a-b) · (a-b)^5 = -(a-b)^{3+1+5} = -(a-b)^9$.
5. (1)若$2^m · 2^n =16$,求$m+n$的值;
(2)已知$a · a^x · a^{2x+1}=a^{29}$,求$x$的值.
(2)已知$a · a^x · a^{2x+1}=a^{29}$,求$x$的值.
答案
解:(1)$\because 2^m · 2^n = 2^{m+n} = 16 = 2^4,\therefore m + n = 4$.
(2)$\because a · a^x · a^{2x+1} = a^{29}$,
$\therefore a^{1+x+2x+1} = a^{29}$(单独的一个字母要看成是本身的1次幂),
$\therefore 3x + 2 = 29$,解得 $x = 9$.
(2)$\because a · a^x · a^{2x+1} = a^{29}$,
$\therefore a^{1+x+2x+1} = a^{29}$(单独的一个字母要看成是本身的1次幂),
$\therefore 3x + 2 = 29$,解得 $x = 9$.
6. [2026 岳阳岳阳楼区月考]若 $a^m = 6, a^n = 3$，则 $a^{m+n}$ 的值为（
A.9
B.18
C.2
D.3
B
）A.9
B.18
C.2
D.3
答案
B $a^{m+n} = a^m · a^n = 6 × 3 = 18$.
7. [2026 厦门外国语学校月考] 若$2^x = 5$，则$2^{x+3}$的值为（
A.8
B.15
C.40
D.125
C
）A.8
B.15
C.40
D.125
答案
C $\because 2^x = 5,\therefore 2^{x+3} = 2^x · 2^3 = 5 × 8 = 40$.
8. [2025周口淮阳区月考] 电子文件的大小常用B，KB，MB，GB等作为单位，其中$1\ \mathrm{GB}=2^{10}\ \mathrm{MB}$，$1\ \mathrm{MB}=2^{10}\ \mathrm{KB}$，$1\ \mathrm{KB}=2^{10}\ \mathrm{B}$. 某视频文件的大小约为1 GB，则1 GB等于
$2^{30}$
B.答案
$2^{30}$ 由题意,得1 GB = $1 × 2^{10} × 2^{10} × 2^{10}$ B = $2^{10+10+10}$ B = $2^{30}$ B.
9. 若$2^{2x+2} - 2^{2x+1} = 32$,求$x$的值.
答案
解:$\because 2^{2x+2} - 2^{2x+1} = 32$,
$\therefore 2 × 2^{2x+1} - 2^{2x+1} = 32,\therefore 2^{2x+1} = 2^5$,
$\therefore 2x + 1 = 5$,解得 $x = 2$.
$\therefore 2 × 2^{2x+1} - 2^{2x+1} = 32,\therefore 2^{2x+1} = 2^5$,
$\therefore 2x + 1 = 5$,解得 $x = 2$.
10. 若$3^x = 2, 3^y = 10, 3^n = 20$, 求 $x, y, n$ 之间满足的数量关系.
答案
解:$\because 3^x = 2, 3^y = 10, \therefore 3^x × 3^y = 2 × 10,\therefore 3^{x+y} = 20$.
又$\because 3^n = 20,\therefore 3^{x+y} = 3^n,\therefore n = x + y$.
又$\because 3^n = 20,\therefore 3^{x+y} = 3^n,\therefore n = x + y$.
