Equations » Logarithmic equations

ln(x) = ^elog(x) with e = 2.71828182846...

Check logarithm rules if needed too.

Example 1

³`log`(`x` – 2)	= 1 + 4 · ³`log`(2)
³`log`(`x` – 2)	= ³`log`(3) + ³`log`(2⁴)
³`log`(`x` – 2)	= ³`log`(3) + ³`log`(16)
³`log`(`x` – 2)	= ³`log`(3 · 16)
³`log`(`x` – 2)	= ³`log`(48)
`x` – 2	= 48
`x`	= 50

Example 2

1 + 2 · ²`log`(`x`)	= ²`log`(5`x` + 3)
²`log`(2) + ²`log`(`x`²)	= ²`log`(5`x` + 3)
²`log`(2`x`²)	= ²`log`(5`x` + 3)
2`x`²	= 5`x` + 3
2`x`² – 5`x` – 3	= 0
`D` = (–5)² – 4 · 2 · –3 = 49

`x` = –(–5) + 2 · 2	or `x` = –(–5) – 2 · 2
`x` = 3	or `x` = –12

Example 3

^0.5`log`(`x` + 6) + 2 · ²`log`(`x`)	= 0
–²`log`(`x` + 6) + ²`log`(`x`²)	= 0
–²`log`(`x` + 6)	= –²`log`(`x`²)
`x` + 6	= `x`²
–`x`² + `x` + 6	= 0
`x`² – `x` – 6	= 0
(`x` + 2)(`x` – 3)	= 0
`x` = –2 or `x` =	3

As x = –2 is not a solution to the original equation, the correct solution is only x = 3.

Example 4
(³log(x) )² – ³log(x) = 0
Substitute: p = ³log(x)

`p`² –	`p` = 0
`p`(`p` –	1) = 0
`p` = 0	or `p` = 1
³`log`(`x`) = 0	or ³`log`(x) = 1
`x` = 1	or `x` = 3

Example 5

⁴`log`(`x`)	= ²`log`(`x` – 2)
²`log`(`x`)²`log`(4)	= ²`log`(`x` – 2)
²`log`(`x`)2	= ²`log`(`x` – 2)
²`log`(`x`)	= 2 · ²`log`(`x` – 2)
²`log`(`x`)	= ²`log`( (`x` – 2)² )
`x`	= (`x` – 2)²
`x`	= (`x` – 2)(`x` – 2)
`x`	= `x`² – 4`x` + 4
0	= `x`² – 5`x` + 4
0	= (`x` – 1)(`x` – 4)
	`x` = 1 or `x` = 4

As x = 1 is not a solution to the original equation, the correct solution is only x = 4.