SOLUTION: CAN SOMEONE, ANYONE PLEASE PLEASE HELP ME!!! I really need help. What is the value of X in the equation, log{{{2}}}(2x+8) - log{{{2}}}(2x^2+21x+61) = -3
Question 20718: CAN SOMEONE, ANYONE PLEASE PLEASE HELP ME!!! I really need help. What is the value of X in the equation, log(2x+8) - log(2x^2+21x+61) = -3 Found 2 solutions by venugopalramana, askmemath:Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! i think you mean logs are to base 2..taking that and using the formula,
log x to base y = log x/log y
so we get
log(2x+8)/log 2+log (2x^2+21x+61)/log 2=-3
multiplying through out with log 2
log(2x+8)+log(2x^2+21x+61)=-3log 2=log 2^(-3)==log(1/2^3)=log(1/8)
log(2x+8)(2x^2+21x+61)=log(1/8)
(2x+8)(2x^2+21x+61)=(1/8)
i hope now you can continue from here
to simplify and factorise to get the answers
You can put this solution on YOUR website! Did you know that logm - log b can be re-written as
log
So you now get log = -3
-3 can be re-written as -3 x 1
and 1 can re-written as log2
which means you now get log
As it is log on both sides now, you can take away the log.
So you get =
Can you solve it from here?
(