Combine the logs on the left hand side. As a reminder, the difference of logs is the log of the quotient.
log[ (x + 4)/(x + 2) ] = log(x)
Take the antilog of both sides,
[(x + 4)/(x + 2)] = x
Solve for x
x + 4 = x(x + 2)
x + 4 = x^2 + 2x
0 = x^2 + 2x - x - 4
0 = x^2 + x - 4
Use the quadratic formula.
x = [ -1 +/- sqrt(1^2 - 4(1)(-4)) ] / 2
x = [ -1 +/- sqrt(1 + 16) ] / 2
x = [ -1 +/- sqrt(17) ] 2
x = { (-1 + sqrt(17))/2 , (-1 - sqrt(17))/2 }
However, be aware that logs are restricted to their domain, in that we cannot take the log of a negative number.
In our original equation, we have log(x + 4) (which means log must be greater than -4), log(x + 2) (which means log must be greater than -2), and log(x) (which means log must be greater than 0). This means x must be positive, and (-1 - sqrt(17))/2 is definitely not positive. We reject this solution, and our only solution is
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log(x+4)/log(x+2) = log(x)
(x+4)(x+2) = x
expand then use the quadr formula or factor to get the values of x.
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x=2
log(x + 4) - log(x + 2) = log(x)
Combine the logs on the left hand side. As a reminder, the difference of logs is the log of the quotient.
log[ (x + 4)/(x + 2) ] = log(x)
Take the antilog of both sides,
[(x + 4)/(x + 2)] = x
Solve for x
x + 4 = x(x + 2)
x + 4 = x^2 + 2x
0 = x^2 + 2x - x - 4
0 = x^2 + x - 4
Use the quadratic formula.
x = [ -1 +/- sqrt(1^2 - 4(1)(-4)) ] / 2
x = [ -1 +/- sqrt(1 + 16) ] / 2
x = [ -1 +/- sqrt(17) ] 2
x = { (-1 + sqrt(17))/2 , (-1 - sqrt(17))/2 }
However, be aware that logs are restricted to their domain, in that we cannot take the log of a negative number.
In our original equation, we have log(x + 4) (which means log must be greater than -4), log(x + 2) (which means log must be greater than -2), and log(x) (which means log must be greater than 0). This means x must be positive, and (-1 - sqrt(17))/2 is definitely not positive. We reject this solution, and our only solution is
x = [-1 + sqrt(17)]/2
log [ (x + 4) / (x + 2) ] = log x
(x + 4) / (x + 2) = x
x + 4 = x² + 2x
x² + x - 4 = 0
x = [ - 1 ± √ (1 + 16 ) ] / 2
x = [ - 1 ± √ (17) ] / 2
x = [ - 1 + √ (17) ] / 2 is acceptable because > 0
We seem to agree !
logx^4-logx^2 = 4logx-2logx=2logx 2logx = 2 logx = 1 x = 10^1 = 10
Log a/b = Log a - log b
So from your equation, we get
Log (x+4)/(x+2) = log x
Taking antilog on both sides, we get
x+4 = x(x+2)
x^2 + 2x - x - 4=0
x^2 + x - 4 = 0
Solving this equation, we get
x =[ -1 +- sqrt (1-4*1*-4) ] / 2
= (-1 +- sqrt 17)/2
x+4/x+2=x
x^2+2x=x+4
x^2+x-4=0
x=-1+(17)^1/2 /2=1.561553
log5.561553 - log3.561553=1.561553 o.k.
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[-1-(17)^1/2 / 2] NOT acceptable because logarithm is not defined for negative numbers.
log[ (x+4)/(x+2)] = log(x)
(x+4)/(x+2)=x
(x+4)=x(x+2)
x+4=x^2+2x
x^2+2x-x-4=0
x^2+x-4=0
This equation is of form ax^2+bx+c
a = 1 b = 1 c = -4
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[-1 +/-sqrt(1^2-4(1)(-4)]/(2)(1)
discriminant is b^2-4ac =17
x=[-1 +√(17)] / (2)(1)
x=[-1 -√(17)] / (2)(1)
Yes, you're correct.
log(x+2)
Your answer is exactly correct!!!!