SOLUTION: How do I solve algebraically for all values of x:
log(x+3)*(2x+3)+log(x+3)*(x+5)=2
In this case, (x+3) is the base.
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Question 948119: How do I solve algebraically for all values of x:
log(x+3)*(2x+3)+log(x+3)*(x+5)=2
In this case, (x+3) is the base.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
How do I solve algebraically for all values of x:
log(x+3)*(2x+3)+log(x+3)*(x+5)=2
In this case, (x+3) is the base.
***
convert to exponential form: base(x+3) raised to log of number(2)=numberz((2x+3)(x+5))
(x+3)^2=(2x+3)(x+5)
x^2+6x+9=2x^2+13x+15
x^2+7x+6=0
(x+6)(x+1)=0
x=-6 (reject, (x+3)<0
or
x=1
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