Solve for x. log(x) + log(x - 3) = 1 Please explain how you got the answer.
familiar with log rules ?
No I forgot them. That's why I need help with this. Can you help me?
quotient rule\[\large\rm log_b y - \log_b x = \log_b \frac{ x }{ y}\] to condense you can change subtraction to division product rule \[\large\rm log_b x + \log_b y = \log_b( x \times y )\] addition ----> multiplication power rule \[\large\rm log_b x^y = y \log_b x\]
so there is a plus sign which rule you should apply ?
addition ----> multiplication
How do you know which number is b?
yep so change addition to multiplication log is same so you can apply it
b is base
so log is same as log base 10 \[\huge\rm log = \log_{10}\]
So is it log(x*(x-3))?
Or log(x^2-3x)?
yep right!!
So what do you do with the 1?
now you have \[\huge\rm log (x^2 -3x) =1\] next step convert log to exponential form
How do you do that again?
and here is the example |dw:1438129295617:dw|
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