Question

Please help me solve this problem.

Answer 1

\[1/2-\log_{16} (x)=\log_{16} (x+3)\]

Answer 2

put all logs on one side then add them by using log a+ logb = logab then use log a with base b=C then a=b^C so this is how it goes
1/2 = log x + log (x+3) = log x(x+3) = log(x^2 +3x) so then x^2 + 3x = 16^1/2 = 4 so it becomes a quadratic formula x^2 + 3x -4=0 so if you factorise (x-1)(x+4) =0 answers are x=1 and X=-4

Answer 3

why is it log(x^2+3x)? What happened to base 16?

Answer 4

I didnt see the equation button but you must put the base 16 in since they are the same you can add or take away

Answer 5

So... I can solve it without base 16 because they are all under the same base?

Answer 6

Yes that is why I wrote 16^1/2 because base is 16

Answer 7

Oh.. okay. I understand now. thanks. :D

Answer 8

ok going to bed now

Answer 9

Goodnight!

Answer 10

good night