Question

can someone check this out?

Answer 1

What?

Answer 2

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

Answer 3

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

Answer 4

noit is not correct

Answer 5

means solve for x, yes?

Answer 6

yes.

Answer 7

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

Answer 8

first
\[2log_2(x)=log_2(x^2)\]

Answer 9

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

Answer 10

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

Answer 11

\[x^2=4(x+3)\]
\[x^2=4x+12\]
\[x^2-4x-12=0\]

Answer 12

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

Answer 13

hope the steps are clear.

Answer 14

thank you..

Answer 15

welcome!