SOLUTION: log(x^2+7x)=log18
I think I'm supposed to get the log on one side, so would it look like this? log18(x^2+7x)
Do I then write it out like this? log18 x^2 +log18 7x
Algebra.Com
Question 966659: log(x^2+7x)=log18
I think I'm supposed to get the log on one side, so would it look like this? log18(x^2+7x)
Do I then write it out like this? log18 x^2 +log18 7x
Found 2 solutions by Theo, josmiceli:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
log(x^2 + 7x) = log(18) can only be true if x^2 + 7x = 18
so, start with x^2 + 7x = 18.
subtract 18 from both sides of the equation to get x^2 + 7x - 18 = 0
factor this quadratic equation to get (x + 9) * (x - 2) = 0
solve for x to get x = -9 or x = 2.
go back to your original equation to see if these values are good.
they have to be good in the original equation, or they are not good.
your original equation is log(x^2 + 7x) = log(18)
when x = -9, you get log(x^2 + 7x) = log(18) becomes log(81 - 63) = log(18) which becomes log(18) = log(18) which is true, so x = -9 is good.
when x = 2, you get log(x^2 + 7x) = log(18) becomes log(4 + 14) = log(18) which becomes log(18) = log(18) which is true, so x = 2 is also good.
your solution is that x = -9 or x = 2.
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
What you need here is the general rule:
If
Then,
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check answers:
You can check other solution,
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