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log x + log 2x = 3 log 2    find x values

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clara2 | Student, Undergraduate | eNoter

Posted November 7, 2010 at 10:22 AM via web

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log x + log 2x = 3 log 2    find x values

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

Posted November 7, 2010 at 10:23 AM (Answer #1)

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log x + log 2x = 3 log 2

We know that:

log a + log b = log a*b

==> log x + log 2x = log 2x^2

==> log 2x^2 = 3 log 2

Also, we know that:

a log b = log b^a

==> 3 log 2 = log 2^3 = log 8

==> log 2x^2 = log 8

==> 2x^2 = 8

Now divide by 2:

==> x^2 = 4

==> x= 2

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neela | High School Teacher | Valedictorian

Posted November 7, 2010 at 10:32 AM (Answer #2)

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logx+log2x= 3log3.

To solve for x.

Solution:

We use the following property of logarithms:

loga+logb =  logab...(1)

And m*loga = loga^m...(2)

So logx+log2x= 3log2.Using property (1), thegiven  equation  becomes logx*3x = 3log2

log3x^2 = log2^3. Here we used the property (2).

Taking antilog, we get:

3x^2 = 2^3 = 8

x^2 = 8/3.

x = sqrt(8/3). Or  x= -sqrt(8/3), but the negative solution is invalid as logarithms are undefined for negatives .

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shurlo | Student, Grade 9 | eNotes Newbie

Posted November 9, 2010 at 1:44 AM (Answer #3)

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log x+log2x=3log2

log(x*2x)=log2 to the power 3

log2x^2=log8

x^2=8/2

x=+_2

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giorgiana1976 | College Teacher | Valedictorian

Posted November 7, 2010 at 6:05 PM (Answer #4)

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We'll start by imposing constraints of existence of logarithms:

x>0

2x>0

The interval of admissible values of x is (0 , +infinite).

Now, we'll solve the equation. We'll start by applying the power property of logarithms:

3 log 2 = log 2^3 = log 8

Now, we'll subtract log 2x both sides:

log x = log 8 - log 2x

We'll apply the quotient rule of logarithms:

log a - log b = log a/b

log x = log 8/2x

log x = log 4/x

Since the bases are matching, we'll apply the one to one property:

x = 4/x

x^2 = 4

x1 = +sqrt4

x1 = +2

x2 = -2

We'll reject the second solution because it doesn't belong to the interval of admissible values.

The only solution of the equation is x = 2.

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