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11. [2025厦门期末]已知$x + y - 3 = 0$，则$3^x · 3^y$的值是（
A.$9$
B.$27$
C.$\dfrac{1}{2}$
D.$\dfrac{1}{27}$
B
）A.$9$
B.$27$
C.$\dfrac{1}{2}$
D.$\dfrac{1}{27}$
答案
B $\because x + y - 3 = 0,\therefore x + y = 3,\therefore 3^x · 3^y = 3^{x+y}$(整体代入) $= 3^3 = 27$.
12. [2026 盐城阜宁期末] 若$a,b$是正整数，$4^a + 4^a + 4^a + 4^a = 4^b × 4^b × 4^b × 4^b$，则$a$与$b$的关系是（
A.$a = b$
B.$4a = b^4$
C.$a + 1 = b^4$
D.$4b - a = 1$
D
）A.$a = b$
B.$4a = b^4$
C.$a + 1 = b^4$
D.$4b - a = 1$
答案
D $\because 4^a + 4^a + 4^a + 4^a = 4 × 4^a = 4^{a+1},4^b × 4^b × 4^b × 4^b = 4^{b+b+b+b} = 4^{4b},\therefore 4^{a+1} = 4^{4b},\therefore a + 1 = 4b,\therefore 4b - a = 1$.
13. 新情境 [2025沧州模拟] 如图,在甲、乙、丙三只袋中分别装有球29个、29个、5个,先从甲袋中取出$2^x$个球放入乙袋,再从乙袋中取出$(2^x + 2^y)$个球放入丙袋,最后从丙袋中取出$2^y$个球放入甲袋,此时三只袋中球的个数相同,则$2^{x+y}$的值等于 (
A.128
B.64
C.32
D.16
A
)A.128
B.64
C.32
D.16
答案
A 解题思路:先用含$x,y$的式子表示出调整后每个袋子中球的个数,再根据球的总个数求出每个袋子中球的个数,进而列等式求解.
调整后,甲袋中有$(29 - 2^x + 2^y)$个球,乙袋中有$29 + 2^x - (2^x + 2^y) = (29 - 2^y)$(个)球,丙袋中有$5 + 2^x + 2^y - 2^y = (5 + 2^x)$(个)球. $\because$ 共有$29 + 29 + 5 = 63$(个)球,且调整后三只袋中球的个数相同,$\therefore$ 调整后每只袋中有$63 ÷ 3 = 21$(个)球,$\therefore 5 + 2^x = 21,29 - 2^y = 21,\therefore 2^x = 16,2^y = 8,\therefore 2^{x+y} = 2^x · 2^y = 16 × 8 = 128$.
调整后,甲袋中有$(29 - 2^x + 2^y)$个球,乙袋中有$29 + 2^x - (2^x + 2^y) = (29 - 2^y)$(个)球,丙袋中有$5 + 2^x + 2^y - 2^y = (5 + 2^x)$(个)球. $\because$ 共有$29 + 29 + 5 = 63$(个)球,且调整后三只袋中球的个数相同,$\therefore$ 调整后每只袋中有$63 ÷ 3 = 21$(个)球,$\therefore 5 + 2^x = 21,29 - 2^y = 21,\therefore 2^x = 16,2^y = 8,\therefore 2^{x+y} = 2^x · 2^y = 16 × 8 = 128$.
14. (1)若$x^m =5,x^n =7$,求$x^{2m+n}$的值；
(2)若$x^a =3,x^b =4,x^c =5$,求$2x^{a+b+c}$的值.
(2)若$x^a =3,x^b =4,x^c =5$,求$2x^{a+b+c}$的值.
答案
解:(1)$\because x^m =5,x^n =7$,
$\therefore x^{2m+n} = x^{m+m+n} = x^m · x^m · x^n = 5 × 5 × 7 = 175$.
(2)$\because x^a =3,x^b =4,x^c =5$,
$\therefore 2x^{a+b+c} = 2x^a · x^b · x^c = 2 × 3 × 4 × 5 = 120$.
$\therefore x^{2m+n} = x^{m+m+n} = x^m · x^m · x^n = 5 × 5 × 7 = 175$.
(2)$\because x^a =3,x^b =4,x^c =5$,
$\therefore 2x^{a+b+c} = 2x^a · x^b · x^c = 2 × 3 × 4 × 5 = 120$.
15. 规定 $a*b = 3^a × 3^b$，如 $2*3 = 3^2 × 3^3$。
(1) $1*2$ 的值为
(2) 若 $2*(x+1)=81$，求 $x$ 的值。
(1) $1*2$ 的值为
27
；(2) 若 $2*(x+1)=81$，求 $x$ 的值。
答案
(1)27
根据题意,得$1*2 = 3^1 × 3^2 = 3 × 9 = 27$.
(2)$\because 2*(x+1) = 81,\therefore 3^2 × 3^{x+1} = 3^4$,
$\therefore 2 + x + 1 = 4$,解得 $x = 1$.
根据题意,得$1*2 = 3^1 × 3^2 = 3 × 9 = 27$.
(2)$\because 2*(x+1) = 81,\therefore 3^2 × 3^{x+1} = 3^4$,
$\therefore 2 + x + 1 = 4$,解得 $x = 1$.
16. 新趋势·代数推理 已知$3^m + n$能被10整除，试说明$3^{m+4} + n$也能被10整除.
答案
解:$3^{m+4} + n = 3^4 × 3^m + n = 81 × 3^m + n = 80 × 3^m + (3^m + n)$.
$\because 3^m + n$能被10整除,$80 × 3^m$能被10整除,
$\therefore 3^{m+4} + n$也能被10整除.
$\because 3^m + n$能被10整除,$80 × 3^m$能被10整除,
$\therefore 3^{m+4} + n$也能被10整除.
17. 运算能力 观察以下等式：
①$2^{1} + 2^{1} = 2 × 2^{1} = 2^{2}$；
②$2^{2} + 2^{2} = 2 × 2^{2} = 2^{3}$；
③$2^{3} + 2^{3} = 2 × 2^{3} = 2^{4}$；
……
(1)请写出第④个等式：
(2)根据你发现的规律，用含字母$n$的式子表示第$n$个等式：
(3)由(2)中的规律，计算$2^{2026} - 2^{2025}$的值.
①$2^{1} + 2^{1} = 2 × 2^{1} = 2^{2}$；
②$2^{2} + 2^{2} = 2 × 2^{2} = 2^{3}$；
③$2^{3} + 2^{3} = 2 × 2^{3} = 2^{4}$；
……
(1)请写出第④个等式：
$2^4 + 2^4 = 2 × 2^4 = 2^5$
.(2)根据你发现的规律，用含字母$n$的式子表示第$n$个等式：
$2^n + 2^n = 2^{n+1}$
，并说明这个规律的正确性.(3)由(2)中的规律，计算$2^{2026} - 2^{2025}$的值.
答案
(1)$2^4 + 2^4 = 2 × 2^4 = 2^5$.
(2)$2^n + 2^n = 2^{n+1}$
说明这个规律的正确性:$\because 2^n + 2^n = 2 × 2^n = 2^{n+1}$,$\therefore 2^n + 2^n = 2^{n+1}$.
(3)$2^{2026} - 2^{2025} = 2^{2025} + 2^{2025} - 2^{2025} = 2^{2025}$.
(2)$2^n + 2^n = 2^{n+1}$
说明这个规律的正确性:$\because 2^n + 2^n = 2 × 2^n = 2^{n+1}$,$\therefore 2^n + 2^n = 2^{n+1}$.
(3)$2^{2026} - 2^{2025} = 2^{2025} + 2^{2025} - 2^{2025} = 2^{2025}$.
