Question

对数问题求解

Answer 1

可以根据对数定义域，判断c<1.5时无解
。
c>1.5时
方程化为lg（c-1）=0.98*lg（c-0.5）+0.02*lg（c-1.5）
这个方程有唯一实数解。但需要用数值解法，比如二分搜索或者Newton迭代等。或者直接用Matlab或者mathimatic求解。
解为：c
=
2.932031129256685e+14
（手动求解析解基本完蛋）
Matlab二分搜索代码如下：
function
[
y
]
=
calcx2y(
x
)
%
函数方程
y
=
log10(x
-
1)
-
0.98*
log10(x
-
0.5)
-
0.02*
log10(x
-
1.5);
end
主体代码
clear
clc
x1
=
290000000000000;%预估计下界
y1
=
calcx2y(x1)
x2
=
300000000000000;%预估计上界
y2
=
calcx2y(x2)
low
=
x1;
high
=
x2;
th
=
1;%精度需求
while(high-low>th)

mid
=
(high+low)/2;

if(calcx2y(mid)>0)

high
=
mid
-
th;

else

low
=
mid
+
th;

end
end
mid
ymid
=
calcx2y(mid)
c
=
mid;
format
long;
left
=
-0.02*log10((1.5-c)/(0.5-c))
right
=
log10((0.5-c)/(1-c))
error
=
left
-
right
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
我直接把运行结果放在这里了
y1
=

-1.4988e-15
y2
=

1.1102e-15
mid
=

2.9320e+14
ymid
=

1.1102e-15
left
=

2.989415146315196e-17
right
=

7.714619732426290e-16
error
=

-7.415678217794771e-16

Answer 2

解：设(2-2c)/(1-2c)=x

原式可化为：x^50=2x
-
1→x^49
+
x^48
+
....
+
x
-
1=0

根据一元n次方程的解法可求得x，进而可求得c值。