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logarithmsIf log x^3- log 10x = log 10^5 find x
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We'll set constraints of existence of logarithms:
x^3>0 => x>0
10x>0 => x>0
We'll solve the equation adding log 10x both sides:
log x^3 = log 10x + log 10^5
We'll apply the product rule of logarithms:
log x^3 = log x*10^6
Since the bases are matching, we'll apply one to one rule:
x^3 = x*10^6
We'll subtract 10^6x both sides:
x^3 - 10^6*x = 0
We'll factorize by x:
x(x^2 - 10^6) = 0
We'll set each factor as zero:
x = 0
We'll reject this answer since x>0.
x^2 - 10^6 = 0
We'll re-write the difference of squares:
(x - 1000)(x + 1000) = 0
We'll set each factor as zero;
x - 1000 = 0
x = 1000
x + 1000 = 0
x = -1000
The admissible solution of the equation is x = 1000.
Posted by giorgiana1976 on May 5, 2011 at 6:35 AM (Answer #2)
It is given that log x^3- log 10x = log 10^5.
To determine x , use the property : log a - log b = log a/b
log x^3- log 10x = log 10^5
=> log (x^3 / 10x) = 10^5
=> x^2 / 10 = 10^5
=> x^2 = 10^6
=> x = 10^3
=> x = 1000
The solution of the given equation is x = 1000.
Posted by justaguide on May 7, 2011 at 1:45 AM (Answer #3)
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