SOLUTION: Good evening to all I have some difficult to understand the difference between following two logarithmic equations a) log2(x^2 - 4x) =5 with solutions x=-4 V x=8 b) log2(x)

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Good evening to all I have some difficult to understand the difference between following two logarithmic equations a) log2(x^2 - 4x) =5 with solutions x=-4 V x=8 b) log2(x)      Log On


   

Question 1012139: Good evening to all
I have some difficult to understand the difference between following
two logarithmic equations
a) log2(x^2 - 4x) =5 with solutions x=-4 V x=8
b) log2(x) + log2(x-4) = 5 with only one solution x = 8
As I know from log. properties the sum of 2 logarithms is the product of the
arguments, so equation b) gives the product of arguments x^2 - 4x
Where do I wrong?
My thanks in advance

Found 2 solutions by Boreal, josgarithmetic:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
log2(x) + log2(x-4) = 5
log2 (x^2-4x)=5
raise everything to the 2 power
x^2-4x=32
x^2-4x-32=0
(x-8)(x+4)=0
x=8, -4, but can't take a log of a negative number so there is only one solution.
If the log is of (x^2-4x), then x=-4 works.
As soon as you multiply the two together, you have slightly changed the form of the equation. With quadratics and logs, it is necessary to put the root found back into the original equation to see if it works. Often, one does not.

Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
(a)
log%282%2C%28x%5E2-4x%29%29=5
log%282%2C%28x%28x-4%29%29%29=5----correct but unnecessary
2%5E5=x%5E2-4x
x%5E2-4x-2%5E5=0
x%5E2-4x-32=0-----quadratic equation
%28x%2B4%29%28x-8%29=0


(b)
log%282%2Cx%29%2Blog%282%2C%28x-4%29%29=5----this as the original equation is in a different form given. We just do not have log of a negative number, -4. Base two to no power will give negative number.