How do you solve log(x-15)+logx=2?

1 Answer
Mar 7, 2016

S={20}

Explanation:

Use the log property that log(a)+log(b)=log(ab), so

log(x-15)+log(x)=2
log(x(x-15))=2

Since no base was specified we assume the base of the logarithm was 10, take the base-10 exponential of both sides, i.e.:

10^(log(x(x-15)))=10^2

We know that b^(log_b(a))=a, and that 10^2=100 so

(x(x-15))=100

Expand the product and take that 100 to the LHS

x^2-15x-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+-sqrt(625))/(2)=(15+-25)/(2)
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}