Solving A Nice log Equation

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Комментарии
Автор

Actually, in the second method you could use the same substitution as in the first method

NirDagan
Автор

^=read as the power
*=read as square root
As per question

X^(1+logx)=100
(X^1)×(x^logx)=100
According to Lambert w
X^(logx)=x
So,
X×x=100
X^2=100
X=*(100)=±10

ManojkantSamal
Автор

Solution:
x^[1+lg(x)] = 100 |lg() ⟹
[1+lg(x)]*lg(x) = 2 |-2 ⟹
lg²(x)+lg(x)-2 = 0 | with u = lg(x) ⟹
u²+u-2 = 0 |p-q-formula ⟹
u1/2 = -1/2±√(1/4+2) = -1/2±3/2 ⟹
u1 = -1/2+3/2 = 1 and u2 = -1/2-3/2 = -2
1st case: lg(x1) = u1 = 1 |10^() ⟹ x1 = 10
2nd case: lg(x2) = u2 = -2 |10^() ⟹ x2 = 1/100

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