SOLUTION: 2log(x+1)-log(x+2)=log(2x-1)
base log 10
answer=1.3028
Can sir show me the step?
Algebra.Com
Question 457547: 2log(x+1)-log(x+2)=log(2x-1)
base log 10
answer=1.3028
Can sir show me the step?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
2log(x+1)-log(x+2)=log(2x-1)
base log 10
answer=1.3028
..
2log(x+1)-log(x+2)=log(2x-1)
2log(x+1)-log(x+2)-log(2x-1)=0
log(x+1)^2-(log(x+2)+log(2x-1))=0
place under single log
log[(x+1)^2/(x+2)(2x-1)]=0
convert to exponential form (base(10) raised to log of number(0)=number (x+1)^2/(x+2)(2x-1))
10^0=(x+1)^2/(x+2)(2x-1)=1
(x+1)^2=(x+2)(2x-1)
x^2+2x+1=2x^2+3x-2
x^2+x-3=0
solve by quadratic formula as follows:
..
..
a=1, b=1, c=-3
x=[-1ħsqrt(1-4*1*-3)]/2*1
x=(-1ħ√13)/2
x=(-1ħ3.6056)/2
x=-2.3028 (reject, (x+1)>0)
or
x=1.3028 (ans)
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