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What is x if log(x) (1/8) = -3/2.

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clara2 | Student, Undergraduate | eNoter

Posted March 24, 2011 at 11:41 AM via web

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What is x if log(x) (1/8) = -3/2.

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted March 24, 2011 at 11:45 AM (Answer #1)

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Given that log(x) (1/8) = -3/2

We need to find the values of x.

==> Let us simplify.

First we will rewrite (1/8) = (1/2)^3

==> (1/2)^3 = 2^-3

==> log(x) 2^-3 = -3/2

Now we know that log a^b = b*log a

==> log(x) 2^-3 = -3*log(x) 2 = -3/2

We will divide by -3.

==> log(x) 2 = -3/2*-3 = 1/2

Now we will rewrite using exponent form.

==> x^1/2 = 2

We will square both sides.

==> x = 4

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted March 24, 2011 at 12:05 PM (Answer #2)

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We have log(x) (1/8) = -3/2. We have to determine x.

log(x) (1/8) = -3/2

=> (1/8) = x^(-3/2)

=> (1/2)^3 = x^(-3/2)

=> (1/4)^(3/2) = x^(-3/2)

=> 4^(-3/2) = x^(-3/2)

As the exponent are equal we can equate the base.

=> x = 4

The required value of x is x = 4.

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