## anonymous 5 years ago can someone check this out?

1. anonymous

What?

2. anonymous

$2\log_{2}x-\log_{2}(x+3)=2$

3. anonymous

i got $x=2/\log_{2}(x+3)$ i just need to see if its correct?

4. anonymous

noit is not correct

5. anonymous

means solve for x, yes?

6. anonymous

yes.

7. anonymous

ok first combine the left hand side into one logarithm using things like$log(A) + log(B) = log(AB)$

8. anonymous

first $2log_2(x)=log_2(x^2)$

9. anonymous

then $log_2(x^2)-log_2(x+3)=log_2(\frac{x^2}{x+3})$

10. anonymous

so we have $log_2(\frac{x^2}{x+3})=2$ in equivalent exponential form this means $\frac{x^2}{x+3}=2^2=4$

11. anonymous

$x^2=4(x+3)$ $x^2=4x+12$ $x^2-4x-12=0$

12. anonymous

$(x-6)(x+2)=0$ $x=6$or $x=-2$ but we throw out the -2 because you cannot take the log of a negative number. so answer is $x=6$

13. anonymous

hope the steps are clear.

14. anonymous

thank you..

15. anonymous

welcome!