Logarithms » Logarithm rules

You know that 2log(8) = 3, because 23 = 8.
This gives: 22log(8) = 8.

You know that 5log(78 125) = 7, because 57 = 78 125.
This gives: 55log(78 125) = 78 125.

glog(x) = y means gy = x.
Substituting y = glog(x) into gy = x, gives gglog(x) = x.


For g > 0, g ≠ 1, a > 0 and b > 0 the following rules are true:

glog(a) + glog(b) = glog(ab)
glog(a) – glog(b) = glog(ab)

n · glog(a) = glog(an)

glog(a) = plog(a)plog(g) = log(a)log(g)

1glog(a) = – glog(a)


Example 1

Reduce 5 – 3 · 2log(3) to one logarithm.

Answer:
5 – 3 · 2log(3) =
2log(25) – 2log(33) =
2log(2533) =
2log(3227)


Example 2

Solve the equation  1 + 2 · 5log(x) = 7  algebraically.

Answer:

1 + 2 · 5log(x= 7 
2 · 5log(x= 6 
5log(x= 3 
 x = 53
x = 125

Example 3

Solve the equation 2 · 2log(x) + 0,5log(x + 6) = 0 algebraically.

Answer:

2 · 2log(x) + 0,5log(x + 6) = 0
2log(x2) + 2log(x + 6)2log(0,5) = 0
2log(x2) – 2log(x + 6) = 0 
2log(x2= 2log(x + 6)
x2 = x + 6 
x2 – x – 6 = 0 
(x + 2)(x – 3) = 0
x = –2 of x = 3

Because negative numbers within a logarithm do not have an outcome, x = –2 is not a valid solution to the original equation.