《2025年金版教程高中新课程创新导学案高中数学必修第一册人教版》

2.计算：
(1)$(\log_43+\log_83)\frac{\lg2}{\lg3}$；
(2)$(\log_2125+\log_425+\log_85)(\log_52+\log_{25}4+\log_{125}8)$。

答案： 2.解：
(1)原式=$\frac{(\frac{\lg3}{\lg4} + \frac{\lg3}{\lg8})\frac{\lg2}{\lg3}}{\frac{\lg2}{1.8}} = \frac{\lg3}{2\lg2} × \frac{\lg2}{\lg3} + \frac{\lg3}{3\lg2} × \frac{\lg2}{1.8} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$.
(2)解法一：原式=$(\log_2{5^3} + \frac{\log_2{25}}{\log_2{4}} - \frac{\log_8{5}}{\log_8{125}})(\log_2{2} + \log_2{1} - \log_2{8}) =$
$(\frac{3\log_2{5} + 2\log_2{5} - \log_5}{2\log_2{2}} + \frac{3\log_2{2} - 3\log_2{5}}{3\log_2{2}})(\log_2{2} + \frac{2\log_2{2} - 3\log_2{5}}{3\log_2{2}}) =$
$(3 + 1 + \frac{1}{3})\log_2{5} × 3\log_2{5} = 13\log_2{5} × \frac{\log_2{2}}{\log_2{5}} = 13$.
解法二：原式=$(\frac{\lg125}{\lg2} + \frac{\lg25}{\lg4} + \frac{\lg5}{\lg8})(\frac{\lg2}{\lg25} + \frac{\lg4}{\lg25} + \frac{\lg8}{\lg125}) =$
$(\frac{3\lg5 + 2\lg5 - \lg5}{\lg2} + \frac{2\lg5 + 2\lg2 - 3\lg5}{2\lg2} + \frac{3\lg2 - 3\lg5}{3\lg2})(\frac{\lg2 + 2\lg2 + 3\lg2}{\lg5} + \frac{2\lg2 + 2\lg2 + 3\lg2}{3\lg5}) =$
$\frac{13\lg5 × 3\lg2}{3\lg2 × \lg5} = 13$.
解法三：原式=$(\log_2{5^3} + \log_2{5^2} + \log_2{5^{\frac{1}{3}}})(\log_2{2} + \log_2{2^2} + \log_2{2^3}) =$
$(3\log_2{5} + \log_2{5} + \frac{1}{3}\log_2{5})(\log_2{2} + \log_2{2} + \log_2{2}) =$
$(3 + 1 + \frac{1}{3}) × \log_2{5} × 3\log_2{2} = \frac{13}{3} × 3 = 13$.

例3 (1)方程$\log_3(x-1)=\log_3(x+\sqrt3)+\log_3(x-\sqrt3)$的解为(B)
A.$-1,2$
B.$2$
C.$-1$
D.$1,-2$
(2)已知$2^x=3^y=5^z$，且$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$，求$x,y,z$。

答案：
(1)[解析] 由$\log_3{(x + \sqrt{3})} + \log_3{(x - \sqrt{3})} = \log_3{[(x + \sqrt{3})(x - \sqrt{3})]} = \log_3{(x^2 - 3)}$,可得$\log_3{(x - 1)} = \log_3{(x^2 - 3)}$,
所以$\begin{cases}x - 1 = x^2 - 3,\\x + \sqrt{3} ＞ 0,\\x - \sqrt{3} ＞ 0,\\x - 1 ＞ 0.\end{cases}$解得$x = 2$.
[答案] B
(2)[解] 令$2^x = 3^y = 5^z = k(k ＞ 0)$,
$\therefore x = \log_2{k},y = \log_3{k},z = \log_5{k}$,
$\therefore \frac{1}{x} = \log_k{2},\frac{1}{y} = \log_k{3},\frac{1}{z} = \log_k{5}$,
由$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$,
得$\log_k{2} + \log_k{3} + \log_k{5} = \log_k{30} = 1$,
$\therefore k = 30$,
$\therefore x = \log_2{30} = 1 + \log_2{15}$,
$y = \log_3{30} = 1 + \log_3{10}$,
$z = \log_5{30} = 1 + \log_5{6}$.