SOLUTION: I cannot figure out the solution to this problem i have been trying for days assistance would be very much appreciated please 1.) solve for x: log_4(x^2+3x)-log_4(x+5)= -im

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I cannot figure out the solution to this problem i have been trying for days assistance would be very much appreciated please 1.) solve for x: log_4(x^2+3x)-log_4(x+5)= -im      Log On


   

Question 276770: I cannot figure out the solution to this problem i have been trying for days assistance would be very much appreciated please

1.) solve for x: log_4(x^2+3x)-log_4(x+5)=
-im not sure if i divide or subtract or pretty much anything that im suppose to do i cannot figure it out
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
I'm not sure how well you understand logs. You have to
know the general rules, The best rule to remember is
"logs are exponents". The rule that applies here is:
log%28a%2C%28b%2Fc%29%29+=+log%28a%2Cb%29+-+log%28a%2Cc%29
Your problem:
log%284%2C%28x%5E2+%2B+3x%29%29+-+log%284%2C%28x%2B5%29%29
Using the rule in reverse,
log%284%2C%28%28x%5E2+%2B+3x%29%2F%28x+%2B+5%29%29%29
In order to solve for x, you need this expression
to equal some number, for instance, if it equals 3,
log%284%2C%28%28x%5E2+%2B+3x%29%2F%28x+%2B+5%29%29%29+=+3
This is the same as
4%5E3+=+%28x%5E2+%2B+3x%29%2F%28x%2B5%29
64+=+%28x%5E2+%2B+3x%29%2F%28x%2B5%29
64x+%2B+320+=+x%5E2+%2B+3x
x%5E2+-+61x+-+320
This can be solved using quadratic formula