Use the log property that log(a)+log(b)=log(ab)log(a)+log(b)=log(ab), so
log(x-15)+log(x)=2log(x−15)+log(x)=2
log(x(x-15))=2log(x(x−15))=2
Since no base was specified we assume the base of the logarithm was 1010, take the base-10 exponential of both sides, i.e.:
10^(log(x(x-15)))=10^210log(x(x−15))=102
We know that b^(log_b(a))=ablogb(a)=a, and that 10^2=100102=100 so
(x(x-15))=100(x(x−15))=100
Expand the product and take that 100 to the LHS
x^2-15x-100=0x2−15x−100=0
Solve the quadratic in your favorite way, by the quadratic formula for example
x=(15+-sqrt(15^2-4*1*(-100)))/(2*1)=(15+-sqrt(225+400))/(2)x=15±√152−4⋅1⋅(−100)2⋅1=15±√225+4002
x=(15+-sqrt(625))/(2)=(15+-25)/(2)x=15±√6252=15±252
x^'=(15+25)/(2)=(40)/(2)=20
x^'=(15-25)/(2)=(-10)/(2)=-5
However, recall that this was originally a logarithmic formula; we can have anything equal to 0 or a negative number in one!
So we know that
x-15>0
x>0
20-15=5>0 and 20>0 so it's a valid solution, but -5<0 so it isn't one. So the set of solutions is S={20}
