Logarithms » Logarithm rules

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

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

^glog(x) = y means g^y = x.
Substituting y = ^glog(x) into g^y = x, gives g^{^glog(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)

^{^{^1g}}log(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 · ⁵`log`(`x`)	= 7
2 · ⁵`log`(`x`)	= 6
⁵`log`(`x`)	= 3
`x`	= 5³
`x`	= 125

Example 3
Solve the equation 2 · ²log(x) + ^0,5log(x + 6) = 0 algebraically.

Answer:

2 · ²`log`(`x`) + ^0,5`log`(`x` + 6)	= 0
²`log`(`x`²) + ²`log`(`x` + 6)²`log`(0,5)	= 0
²`log`(`x`²) – ²`log`(`x` + 6)	= 0
²`log`(`x`²)	= ²`log`(`x` + 6)
`x`²	= `x` + 6
`x`² – `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.