Expert: Paul Klarreich - 3/30/2008

Question
So I am stumped on how to solve these Logs:
12. log5(x - 1) + log5(x - 2) - log5(x + 6) = 0
13.a) log5(log3x) = 0 *where 3 is subscript
b) log2(log4x) = 1 *where 4 is subscript
14a) 1/2loga(x + 2) + 1/2loga(x - 1) = 2/3loga27 * a is subscript
14b)logb(x - 1) + logb(x + 2) = logb(8 - 2x)

Answer
Questioner:   khaleel
Category:  Advanced Math
Private:  No
 
Subject:  Logarithms
Question:  So I am stumped on how to solve these Logs:
12. log5(x - 1) + log5(x - 2) - log5(x + 6) = 0
13.a) log5(log3x) = 0 *where 3 is subscript
b) log2(log4x) = 1 *where 4 is subscript
14a) 1/2loga(x + 2) + 1/2loga(x - 1) = 2/3loga27 * a is subscript
14b)logb(x - 1) + logb(x + 2) = logb(8 - 2x)
..............................
Hi, Khaleel,

That's a lot of questions, so let me just get you started with a couple of hints:

Use these log properties: [Logb means logarithm in base-b.]
I)   logb(x) + logb(y) = logb(xy)  
II)  logb(x) - logb(y) = logb(x/y)  
III) n logb(x) = logb(x^n), and n may be a fraction.

IV)   logb(x) = n means  x = b^n
IVa)  logb(x) = 0 means  x = 1
IVb)  logb(x) = 1 means  x = b
V)  if logb(x) = logb(y), then  x = y

So you should be able to handle your examples now: I will get you started.

12. log5(x - 1) + log5(x - 2) - log5(x + 6) = 0

log5[ (x - 1)(x - 2)/(x + 6) ] = 0

(x - 1)(x - 2)/(x + 6) ] = 1

etc.
.....................
13.a) log5(log3x) = 0 *where 3 is subscript << NO, it's the base.

log5(log3x) = 0

log3(x) = 1

etc.

b) log2(log4x) = 1

log4(x) = 2

etc.
.......................

14a) 1/2loga(x + 2) + 1/2loga(x - 1) = 2/3loga27

loga(x + 2)^1/2  + loga(x - 1)^1/2 = loga27^2/3

loga[ (x + 2)^1/2(x - 1)^1/2] = loga 9

(x + 2)^1/2(x - 1)^1/2 = 9

etc.
.......................
14b)logb(x - 1) + logb(x + 2) = logb(8 - 2x)

logb[(x - 1)(x + 2)] = logb(8 - 2x)

etc.