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


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.


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:


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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