More Math Help

[Namine]

1. Combine the sum of ln first so:

ln [45(x-4)] = ln (5x-4)

Put everything as powers of e, so you have

e^[ln 45(x-4)] = e^[ln (5x-4)]

That'll cancel out all the ln, so you are only left with

45(x-4) = 5x-4

Just solve for x then.



2. Expand the difference of square first, so:

log [ (x+7)(x-7)] - 1 = log (x+7)

Break up the multiplication inside the log there:

log (x+7) + log (x-7) - 1 = log (x+7)

log (x-7) = 1

Since log10 = 1, x-7 must also equal to 10, so x is 17.




3. Change to base 10 (or use ln, but i hate ln) so it's easier to deal with first.

log(13-x)/log3 - log(x+3)/log3 = 1

log(13-x) - log(x+3) = log3

Again change the difference of log, combining into division for the number inside log.

log[(13-x)/(x+3)] = log3

Since both side are the same, compare the number inside the log function.

(13-x)/(x+3) = 3

And now you can solve for x.