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solve for x.   logx^2 +log4 = logx-log2

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samerrima | Student, Grade 9 | eNoter

Posted June 19, 2010 at 7:49 AM via web

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solve for x.   logx^2 +log4 = logx-log2

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

Posted June 19, 2010 at 7:53 AM (Answer #1)

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logx^2 + log4 = logx-log2

Let us rearrange by moving x terms to the left side and the constant terms to the right side:

logx^2 -logx = -log4 -log2

Now we know that logx^2 = 2logx

==> 2logx - logx = -(log4+log2)

Now we know that logx+logy = logxy

==> logx = -log8

But -logx= logx^-1

==> logx = log8^-1 = log(1/8)

==> x= 1/8

To check:

logx^2 +log4 = logx-log2

log(1/64) + log4 = log(1/8)-log2

log (4/64) = log (1/16)

log(1/26) = log(1/16)

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brianevery | High School Teacher | (Level 2) Adjunct Educator

Posted June 19, 2010 at 9:30 AM (Answer #2)

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Log(x^2)+log4=logx-log2

first lets subtract logx from both sides and log4 from both sides to separate the variables from the constants, this gives us

log(x^2)-logx=-log2-log4

on the left side we can use the rule that loga-logb = log(a/b) to get logx^2-logx = log((x^2)/x) which simplifies to logx

on the right side we will first factor out a -1 to get -log2-log4 = -(log2+log4) and we will use the rule that loga+logb = log(a*b) to get -(log(2*4)) or -(log8) where we will use the rule alogb = logb^a to get -(log8) = log(8^-1) or log(1/8)

Now we have

logx=log(1/8) so therefore x must equal (1/8)

We can check that by substituting (1/8) for x in the original equation and evaluating

log(1/8)^2+log4=log(1/8)-log2   substitute (1/8) for x

log(1/64)+log4=log(1/8)-log2 square 1/8

log((1/64)*4)=log((1/8)/2)  follow log(a*b) and log(a/b) rule

log(4/64)=log((1/8)/(2/1))  simplify

log(1/16)=log((1/8)*(1/2)) invert and mulitply, simplify

log(1/16)=log(1/16)   checks

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neela | High School Teacher | Valedictorian

Posted June 19, 2010 at 11:10 AM (Answer #3)

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log2 +log4 =logx-log2. To find x.

Solution:

We say log of logaritm to base 10, just as ln to natural logarithm to base e.

We use the logarithm rule: loga+logb = log(ab) and log a-log b = log(a/b)

logx^2+log4=logx-log2. Implies:

log (x^2 * 4) = log(x/2). Take antilogarithms

4x^2=x/2. Multiply by2.

8x^2 = x.

8x^2-x = 0. Or

x(8x-1) = 0.

x = 0 . Or 8x-1 = 0. Or

x = 1/8.

 

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giorgiana1976 | College Teacher | Valedictorian

Posted June 20, 2010 at 1:37 AM (Answer #4)

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Before solving the equation, let's impose the constraint of existence of logarithms, which is x>0.

Now, we'll apply the product rule of logarithms, to the left side of equality, so that logx^2 +log4 = log (4*x^2).

To the right side of equality, we'll apply the quotient rule of logarithms, so that logx-log2 = log (x/2).

We'll re-write the equation, according to the new results:

log (4*x^2) = log (x/2)

We'll apply now the one to one rule:

4*x^2 = x/2

We'll multiply by 2 both sides:

8x^2 = x

We'll move all terms to the left side:

8x^2 - x = 0

We'll factorize and we'll get:

x(8x-1)=0

We'll put each factor equal to zero:

x=0, which is impossible because x>0!

8x-1 = 0

We'll add 1 both sides:

8x = 1

x = 1/8

Because 1/8>0, the only solution of the equation is x = 1/8.

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