I cannot remember how to do this . any help would be appeciated. logx - log(x+3) = -1 also 3^(x^2+4x)=27 Please help me! Thank You!
Follow Math Help Forum on Facebook and Google+
Originally Posted by hdrag02 I cannot remember how to do this . any help would be appeciated. logx - log(x+3) = -1 also 3^(x^2+4x)=27 Please help me! Thank You! $\displaystyle \log x- \log(x+3)=-1$ $\displaystyle \log \frac{x}{x+3}=-1$ $\displaystyle 10^{-1}=\frac{x}{x+3}$ Can you finish?? $\displaystyle 3^{x^2+4x}=27$ $\displaystyle 3^{x^2+4x}=3^3$ $\displaystyle x^2+4x=3$ Can you finish?
I think i have the right answer, but it seems weird. Heidi
Originally Posted by hdrag02 I think i have the right answer, but it seems weird. Heidi What is it and we'll see how weird it is?
is the answer 1/3? for the first problem and -2 + or - the sq root of 7 for the other? I just feel like an idiot. Heidi
Originally Posted by hdrag02 is the answer 1/3? for the first problem and -2 + or - the sq root of 7 for the other? I just feel like an idiot. Heidi Don't know why you feel that way. Your answers are correct!!
Thank You soooooo Much! I really appreciated your help. I will be back! Heidi
Is there any way you can send me your answer? I was told I had the wrong answer. My Professor had x^2+4x=-3. I am confused. could you explain this to me? Thank you.Heidi
View Tag Cloud