log(1+x)=2logx.solve it
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answer these questions is 1
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log(x+1) = 2logx
now
log(x+1) = logx^2
so compare both side
x+1= x^2
x^2 - x - 1 =0
then
x = 1+√5/2
&
x = 1 - √5/2 but this value is not possible because it is negative
also we know that
in logx
only define when
x > 0
so here only
x =1+√5/2 is the solution of the equation.
now
log(x+1) = logx^2
so compare both side
x+1= x^2
x^2 - x - 1 =0
then
x = 1+√5/2
&
x = 1 - √5/2 but this value is not possible because it is negative
also we know that
in logx
only define when
x > 0
so here only
x =1+√5/2 is the solution of the equation.
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