Let's have a look at one log problem:
Question #1:
log5 x = 1 + 2 logx 5
Find value of x
Solution:
log5 x = 1 + 2 logx 5
log5 x = 1 + 2/log5 x ( Note: change of base in formula : loga b = 1/logb a )
(log5 x)^2 = log5 x + 2 (Note: multiply both side with log5 x to eliminate the denominator)
(log5 x)^2 - log5 x - 2 = 0 ( Note: move everything to left hand side)
(log5 x - 2) ( log5 x + 1) = 0 ( Note: factorise)
When log5 x = 2
x = 5^2
x=25
When log5 x = -1
x = 5^-1
x = 1/5
Hence, x = 1/5 and 25
If you are not sure about your answers, replace the value of x with 1/5 and 25 in the equation and see if they fit in.
log5 25 = 1 + 2 log25 5
2 = 1 + 2(0.5)
= 1 + 1
log5 1/5 = 1 + 2 log1/5 5
-1 = 1 + 2(-1)
= 1 - 2
= -1
Question #2 :
log5 pq = 2 + 3 log5 p - log5 q
Express p in term of q
Solution: The steps below are elaborated to allow for detailed explanations - refer the note on each respective lines. Some obvious lines can be omitted in your answer.
log5 pq = 2 + 3 log5 p - log5 q
log5 p + log5 q = log5 5^2 + 3 log5p - log5 q ( Note 1: use the log law loga MN = loga M + loga N)
(Note 2: Remember: loga M^b = b loga M)
log5 p - 3 log5 p = log5 5^2 - log5 q - log5 q (Note 3: move all p to the left side and all q to the right side.)
-2 log5 p = log5 5^2 - 2log5 q
2log5 p = -log5 5^2 + 2 log5 q ( Note 4 : change negative to positive)
2 log5 p = - 2 log5 5 + 2 log5 q
log5 p = - log5 5 + log5 q ( Note 5: divide with 2)
lo5 5 p = log5 (q/5)
p = q/5
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