1、化简: $${(27^{1over log_23} + 5^{log_{25}49})(81^{1over log_49} - 8^{log_49})over 3+5^{1over log_{16}25} cdot 5^{log_53}}.$$ 解答:
原式 = ${left(27^{log_32} + 5^{log_57} ight)cdotleft(81^{log_94} - 8^{log_23} ight) over 3 + 5^{log_{25}16}cdot 5^{log_53}} = {(8 + 7)cdot(16 - 27)over 3 + 4cdot 3}= -11$.
2、解方程 $log_5(4^x + 144) - 4log_52 = 1 + log_5(2^{x - 2} + 1)$.
解答: $$ecause log_5left(4^x + 144 ight) - log_516 = 1 + log_5left(2^{x - 2} + 1 ight)Rightarrow log_5{4^x + 144 over 16} = log_5left[5cdotleft(2^{x - 2} + 1 ight) ight]$$ $$Rightarrow {4^x + 144 over 16} = 5cdotleft(2^{x - 2} + 1 ight) Rightarrow 2^{2x} + 144 = 80cdotleft({1over4}cdot 2^x + 1 ight)$$ $$Rightarrow left(2^x ight)^2 - 20cdot 2^x + 64 = 0 Rightarrow 2^x = 4, 16$$ $$ herefore x_1 = 2, x_2 = 4.$$ 经检验, $x_1 = 2$, $x_2 = 4$ 是原方程的解.
3、设 $d > 0$, $d e 1$, $y = d^{1over 1-log_dx}$, $z = d^{1over 1-log_dy}$, 求证: $$x = d^{1over 1 - log_dz}.$$ 解答: $$ecause y = d^{1over 1 - log_dx}, z = d^{1over 1 - log_dy} Rightarrow log_dy = {1over 1-log_dx},$$ $$log_dz = {1over 1-log_dy} Rightarrow log_dz = {1over 1- {1over 1- log_dx}} = {1 - log_dx over -log_dx} Rightarrow left(1 - log_dz ight)cdotlog_dx = 1,$$ $$Rightarrow log_dx = {1over 1 - log_dz} Rightarrow x = d^{1over 1 - log_dz}.$$
4、已知 $a, b, c$ 为不等于1的正数, 且 $abc e 1$, $log_a10 + log_b10 + log_c10 = log_{abc}10$. 求证: $a, b, c$ 中有两个数之积为1.
解答: $$ecause {1over lg a} + {1over lg b} + {1over lg c} = {1over lg(abc)} = {1over lg a + lg b + lg c}$$ $$Rightarrow {lg a + lg b over lg a cdot lg b} + {lg a + lg b over left(lg a + lg b + lg c ight)cdot lg c} = 0$$ $$Rightarrow left(lg a + lg b ight)cdot {left(lg a + lg b + lg c ight)lg c + lg a cdot lg b over lg a cdot lg b cdot lg c cdot left(lg a + lg b + lg c ight)} = 0$$ $$Rightarrow {left(lg a + lg b ight)cdotleft(lg b + lg c ight)cdotleft(lg c + lg a ight)over lg a cdot lg b cdot lg ccdotleft(lg a + lg b + lg c ight)} = 0$$ $$Rightarrow left(lg a + lg b ight)cdotleft(lg b + lg c ight)cdotleft(lg c + lg a ight) = 0$$ $$Rightarrow lg ab cdot lg bc cdot lg ca = 0.$$ 即 $ab = 1$ 或 $bc = 1$ 或 $ca = 1$.
5、$x$ 是五位数, $p, q$ 是正整数, 且 $p > q$, 若 $lg x^p$ 和 $lg x^q$ 的尾数相等, 求 $x$ 的值.
解答:
令 $n = lg x^p - lg x^q = lg x^{p - q}in mathbf{N^{*}}$.
$Rightarrow x^{p - q} = 10^n = 2^n cdot 5^n$.
令 $x = 2^acdot 2^b$, $a, binmathbf{N^*}$, $$Rightarrow 2^{acdot(p - q)}cdot 5^{bcdot(p - q)} = 2^n cdot 5^n Rightarrow acdot(p - q) = bcdot(p - q) = n$$ $$Rightarrow a = b Rightarrow x = 10^a Rightarrow a = 4 Rightarrow x = 10000.$$
6、设 $p, q$ 是整数, 且 $1 < q < p < 10$, $displaystyle{qover p}$ 是既约分数, 若 $lgdisplaystyle{1over q}$ 的尾数大于 $lg p^2$ 的尾数, 求 $displaystyle{qover p}$.
解答: $$ecause 1 < p < 10 Rightarrow 0 < lg p < 1$$ $$ecause 1 < q < 10 Rightarrow {1over 10} < {1over q} < 1 Rightarrow -1 < lg{1over q} < 0 Rightarrow -1 < - lg q < 0 Rightarrow 0 < 1 - lg q < 1 Rightarrow lg {1over q} = -lg q = -1 + left(1 - lg q ight).$$
a. 当 $0 < lg p < displaystyle{1over2}$ 时, $$Rightarrow 0 < 2lg p < 1Rightarrow 1 - lg q > 2lg pRightarrow lg p^2q < 1 Rightarrow p^2q < 10.$$ 而 $2^2 imes 3 = 12 > 10Rightarrow$ 无解.
b. 当 $displaystyle{1over2} le lg p < 1$ 时, $$Rightarrow sqrt{10} le p < 10 Rightarrow 0 le 2lg p - 1 < 1 Rightarrow 1 - lg q > 2lg p - 1 Rightarrow lg p^2q < 2 Rightarrow p^2q < 100.$$ 当 $(p, q) = 1$, $1 < q < p < 10$ 且 $sqrt{10} le p < 10$ 时, 满足题意的解为$$(p, q) = (4,3), (5, 2), (5, 3), (7, 2).$$ 综上, $displaystyle{q over p} = {3over4}, {2over5}, {3 over5}, {2over7}$.
7、已知 $lg x, lgdisplaystyle{100over x}$ 的首数分别是 $a$ 和 $b$, 求 $3a^2 - 4b^2$ 的最大值.
解答:
设 $lg x = a + alpha$, $ainmathbf{Z}$, $0 le alpha < 1$, 则 $$lg{100over x} = 2 - lg x = 2 - a - alpha = 1 - a + 1 - alpha.$$ a. 若 $alpha = 0$, 则 $b = 2 - a$. $$3a^2 - 4b^2 = 3a^2 - 4(2-a)^2 = -a^2 + 16a - 16 = -left( a^2 - 16a + 64 ight) + 48$$ $$= -(a - 8)^2 + 48 le 48.$$ b. 若 $0 < alpha < 1$, 则 $b = 1-a$. $$3a^2 - 4b^2 = 3a^2 - 4(1-a)^2 = -a^2 + 8a - 4 = -left( a^2 - 8a + 16 ight) + 12$$ $$= -(a - 4)^2 + 12 le 12.$$ 综上, $left(3a^2 - 4b^2 ight)_ ext{max} = 48$, 此时 $x = 10^8$.
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