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Actually, in the second method you could use the same substitution as in the first method
^=read as the power *=read as square root As per question X^(1+logx)=100(X^1)×(x^logx)=100According to Lambert w X^(logx)=xSo,X×x=100X^2=100X=*(100)=±10
You can discard the -10 solution because you cannot take the log of a negative
x = { 10 , 1/100 }
👍😀✌️👏😎👏✌️😀👍
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 = -21st case: lg(x1) = u1 = 1 |10^() ⟹ x1 = 102nd case: lg(x2) = u2 = -2 |10^() ⟹ x2 = 1/100
Actually, in the second method you could use the same substitution as in the first method
^=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
You can discard the -10 solution because you cannot take the log of a negative
x = { 10 , 1/100 }
👍😀✌️👏😎👏✌️😀👍
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