You can put this solution on YOUR website!
Initially, we can write this as
right...you can do this to the base 10 (or whatever) or better, to base 2. I shall do both, but each version will just say "log", but the base is implied:
To base 10, eqn1 becomes:
log13 - xlog2 = (2x+4)log2
log13 - xlog2 = 2xlog2 + 4log2
2xlog2 + xlog2 = log13 - 4log2
3xlog2 = log13 - 4log2
x(3log2) = log13 - 4log2
--> x =
which is your quoted answer.
However, to the base2, eqn1 becomes:
log13 - x = 2x+4
3x = log13 - 4
--> x =
which is a lot simpler to work with and see, but your calculator doesn't store logs to the base2, so you would have to convert log13 to base10 etc. If you knew that log13 (to the base2) was
(to the base 10), then my second answer does convert to your answer by just manipulating the fraction.