Question

what is the solution set of the equation
log2x+log2(x-3)=2

Answer 1

\(\log (2x) + \log(2(x-3)) = 2\)
like this?

Answer 2

is the question using

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

Answer 3

it is kind of like log (subscript) 2

Answer 4

I think campbell_st has it right.

Answer 5

yes campbell_st

Answer 6

how do i solve it?

Answer 7

ok... so simplify the logs 1st

adding logs means they can be written as 1 and mean multiplying so you have

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

or

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

now raise each part as a power of 2... to remove the logs

\[2^{\log_{2}(x^2 - 3x)} = 2^2...or..... x^2 - 3x = 4\]

so you need to solve the quadratic

\[x^2 - 3x - 4 = 0\]

Answer 8

hope it helps

Answer 9

thanks it does!

Answer 10

so is the solution set (-1,4)

Answer 11

well they are the solutions to the quadratic...

you can check by substituting into the original equation...

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

when x = -1

you get

\[\log_{2}(1 - -3) = \log_{2}(4) = \log_{2}(2^2) = 2\]

so that works

and x = 4

\[\log_{2}(4^2 - 3 \times 4) = \log_{2}(4) .... \]

and you can see the substitution is going the same way as -1 ... so I'd say it works....