Question 493606: Hi I need help with this question:
Evaluate log to the base 2,(2x+1) - log to the base 4,x^2= 1/log to the base 3,2.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Evaluate log to the base 2,(2x+1) - log to the base 4,x^2= 1/log to the base 3,2
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log2(2x+1)-log4(x^2)=1/log3(2)
Convert to log base 10
log(2x+1)/log2-log(x^2)/log4=log(3)/log(2)
log(2x+1)/log2-log(x^2)/log4-log(3)/log(2)=0
log(2x+1)/log(2)-log(x^2)/2log(2)-log(3)/log(2)=0
LCD=2log(2)
2log(2x+1)-log(x^2)-2log(3)=0
place under single log
log[((2x+1)^2)/(x^2)(3^2)]=0
..
Convert to exponential form: base(10) raised to log of number(0)=number[((2x+1)^2)/(x^2)(3^2)]
..
10^0=1=(4x^2+4x+1)/9x^2
(4x^2+4x+1)=9x^2
5x^2-4x-1=0
factor
(5x+1)(x-1)=0
x=-1/5
or
x=1
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