Question 174943
log_3(s) + log_3(s-2)= 1
:
Adding logs means multiply, so we can write it:
log_3(s(s-2)) = 1
:
log_3(s^2-2s) = 1
:
Write the exponent equiv:
{{{3^1}}} = (s^2 + 2s)
or
s^2 - 2s = 3
:
s^2 - 2s - 3 = 0
Factors to
(s-3)(s+1) = 0
:
s = -1; not a solution, s cannot be negative, (cannot have a log of a neg no.)
:
s = 3; is the solution
:
:
Check solution in original problem
log_3(3) + log_3(3-2)= 1
log base 3 of 3 = 1, log base 3 of 1 = 0, so we have:
1 + 0 = 1