1-10对数lg:lg1= 0, lg2 = 0.3010, lg3 = 0.4771, lg4 = 0.6021, lg5 = 0.6990, lg...
注意到依AM-GM不等式,有lg2lg5≤((lg2+lg5)/2)∧2=1/4 故只需证ln2>1/2 又熟知lnx<x-...
=lg5⋅ ( (lg2+lg5) )+lg20+ (2lg5) (lg2)⋅ (2lg2) (lg3)⋅ (2lg3) (lg5) =2+8 =10. 综上所述,结论是:10,结果一 题目 计算${\left(lg5\right)}^{2}+lg2\times lg5+lg20+{\log }_{2}25\times {\log }_{3}4\times {\log }_{5}9$. 答案 (g5)+...
解析 $\left ( {lg5} \right )^{2}+lg2\times lg5+lg2=lg5* ( (lg5+lg2) )+lg2=lg5\times lg\left ( {5\times 2} \right )+lg2=lg5* lg10+lg2=lg5\times 1+lg2=lg5+lg2=lg\left ( {5\times 2} \right )$=lg10=1. 综上所述,答案:1...
$lg2\cdot lg50+\left ( {lg2} \right )^{2}+lg25=lg2⋅ ( (1+lg5) )+ ( (lg2) )^2+2lg5=lg2+lg2⋅ lg5+ ( (lg2) )^2+2lg5=lg2\cdot \left ( {lg2+lg5} \right )+lg2+2lg5=2lg2+2lg5=2\left ( {lg2+lg5} \right )=2\times 1=2$, 综上所述,结论:$lg2\cdot ...
计算:lg ( (lg2+lg5) )=_ _ _ _ _ _ _ _ _ _ .相关知识点: 试题来源: 解析 $\because lg\left ( {lg2+lg5} \right )=lg\left ( {lg\left ( {2\times 5} \right )} \right )=lg ( (lg10) )=lg1=0$ 综上所述填$0$...
∵ lg10=lg ( (2* 5) )=lg2+lg5 ∴ 1=m+lg5 ∴ lg5=1-m 【答案】 C结果一 题目 已知,,则A.n+mB. n-mC. 2-2n+mD. 2+2n+m 答案 由于又由已知,,则故答案为 C.利用换底公式将lg75用lg2与lg3表示出来,再换成用字母n,m表示即可得. 结果二 题目 若lg5=m,则lg2=A、...
$=lg2+lg5=lg10=1$. 综上所述,答案:$1$ (3)由(log)_2 ( (lgx) )=1,可得$lgx=2$, $\therefore x={10}^{2}=100$. 综上所述,答案:$100$ (4)$\left ( {{log}_{3}2+{log}_{9}2} \right )\cdot \left ( {{log}_{4}3+{log}_{8}3} \right )$ $=\left ( {{log}...
既然挂了高中数学的标签lg2⋅lg5≤(lg2+lg52)2=14=ln(e12)12<ln212 ...