solve for x. logx^2 +log4 = logx-log2
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calendarEducator since 2010
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starTop subject is Math
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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calendarEducator since 2008
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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)
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.
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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