solve for x. logx^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
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(1/16)=log((1/8)*(1/2)) invert and mulitply, simplify
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
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:
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.
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)
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.