Hmm, interesting. I'll try to see what method they used. Apparently there is more than one solution to this thing, since if you check by plugging in the answers we got, you would find that they work out as well (by the way, y = 10 if you do it my way, not 30)
what would happen if we started working on the logs equation first?
strange...
I was just thinking about dropping the logs, I don't think we can do that 'easily' in this case. On the RHS of the second equation provided in the question, "x+4" is not in brackets, :. I don't think the log is associated with that so we probably cannot drop the logs like we did...
Hello,
one additional remark to the first problem:
16 = 2^4 . So if you know that the sum x+4 is a logarithm to the base 2 you'll get from
2^(x+4) = 2^x * 2^4 = 2^x * 16
to the last problem:
(4th row of your transformations. You have forgotten to use brackets)
(log_{3}x)^2 - log_{3}x = log_{3}9 Use substitution: log_{3}x = y. Then you have:
y^2 - y - 2 = 0 (log_{3}9 = 2) Solve for y. You'll get y = 2 or y = -1
Now re-substitute:
log_{3}x = 2 ==> x = 9
log_{3}x = -1 ==> x = 3^(-1) = 1/3
Thanks. looks like I have to keep my eyes open for these substitutions, otherwise I'd just solve like I did and get only one ans.to the last problem:
(4th row of your transformations. You have forgotten to use brackets)
(log_{3}x)^2 - log_{3}x = log_{3}9 Use substitution: log_{3}x = y. Then you have:
y^2 - y - 2 = 0 (log_{3}9 = 2) Solve for y. You'll get y = 2 or y = -1
Now re-substitute:
log_{3}x = 2 ==> x = 9
log_{3}x = -1 ==> x = 3^(-1) = 1/3