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# logarithmsIf log x^3- log 10x = log 10^5   find x

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logarithms

If log x^3- log 10x = log 10^5   find x

Posted by kamused on May 5, 2011 at 1:41 AM via web and tagged with algebra1, discussion

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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)

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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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