SOLUTION: please help me solve this. thank you so much.logs confuse me so much log(w^2+1)-log(w-2)=1 thank you so much and God bless.

Question 94597: please help me solve this. thank you so much.logs confuse me so much
log(w^2+1)-log(w-2)=1
thank you so much and God bless.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Logs are "exponents" or "powers" and there are several rules to
keep in mind: log(AB)=logA + logB
log(A/B) = logA -logB
logA^n = nlogA
------------------
Your Problem:
log(w^2+1)-log(w-2)=1
log[(x^2+1)/(w-2)] = 1
---------
So, here you are told "the exponent is 1".
log notation means "the base is 10"
------------
EQUATION:
10^1 = [(x^2+1)/(w-2)]
Cross multiplying you get:
x^2+1 = 10(x-2)
x^2-10x+21=0
(x-7)(x-3)=0
x=7 or x=3
--------
Checking the "answers"
If x=7: log (50) - log (5) = 1
log(50/5) = 1
log(10) = 1
Correct, so x=7 is a good solution;
If x=3: log(10)-log(1) = 1
1-0 = 1
Correct, so x= 3 is also a good solution.
===============
Cheers,
Stan H.

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