You can put this solution on YOUR website! log(3x+7)+log(x-2)=1
log{(3x+7)*(x-2)}=1...SINCE LOG(XY)=LOG(X)+LOG(Y)..NOW TAKING ANTILOGS..WE GET
(3x+7)(x-2)=10....SINCE LOG(10)=1
3X^2-6X+7X-14-10=0
3X^2+X-24=0
3X^2+9X-8X-24=0
3X(X+3)-8(X+3)=0
(3X-8)(X+3)=0
3X-8=0...OR...X=8/3....
X+3=0..OR...X=-3...BUT THIS LEADS TO LOG OF NEGATIVE NUMBERS WHICH DO NOT EXIST.HENCE
X=8/3 IS THE SOLUTION
You can put this solution on YOUR website! Solve for x:
log(3x+7)+log(x-2)=1 ----(1)
log[(3x+7)(x-2)] =1 using log(a) + log(b) = log(ab) all to the same base 10
log[3x(x-2)+7(x-2)] = 1
log[3x^2-6x+7x-14]=1
log(3x^2+x-14)=1
(3x^2+x-14)= (10)^1 [ using log(N) to the base b = p implying N = (b)^p]
(3x^2+x-14)= (10)
3x^2+x-14-10= 0
3x^2+x-24=0
3x^2+[(9x)+(-8x)]-24=0
[splitting the middle term x as the sum of two terms (9x) and (-8x) so that
(9x)X(-8x) = - 72x^2 which is the product of the square term (3x^2) and the constant term (-24)]
(3x^2+9x)-8x-24=0 (by additive associativity)
3x(x+3)-8(x+3)=0
3xp-8p=0 where p = (x+3)
p(3x-8) = 0
(x+3)(3x-8) = 0
(x+3) = 0 implies x=-3 and (3x-8) = 0 implies 3x= 8 giving x = 8/3
Answer: x=-3 and x=8/3
Verification:Putting x=3 in log(3x+7)of (1)
you observe that log(3x+7) = log[3X(-3)+7]= log(-9+2) = log(-2) whcih is not defined. (as log(N) is defined only for N strictly positive)
Therefore x= -3 DOES NOT HOLD in the given equation.
Putting x=8/3 in log(3x+7)+log(x-2)=1 ----(1)
LHS= log(3x+7)+log(x-2)=1 ----(1)
=log[3X(8/3)+7]+log(8/3-2)
=log(8+7)+log[(8-6)/3]
=log15+ log(2/3)
=log[15X(2/3)]
=log5X2)
=log10
=1
(as log of anything to the same base is 1)
Hence x = 8/3 holds
Note: In splitting the middle term x as the sum of two terms (9x) and (-8x) so that (9x)X(-8x) = - 72x^2 which is the product of the square term (3x^2) and the constant term (-24)
How do we hit upon this (+9) and (-8)?
Answer: Multiply the square term (3x^2) and the constant term (-24)
You get -72x^2
72 = 1X2X2X2X3X3
Since 1 multiplied by anything is the samething we may leave out the factor 1 if there are atleast two factors other than 1.
Form two groups of the factors without leaving any factor with an eye on the sign of the middle term and the sign of the product.
Here the mid term is positive and the product is negative.
This means the two parts should be such that the larger numerical one positive and the other negative so that postive larger added to negative smaller is positive and the positive multiplied by negative is negative. So when we try out combinations we hit upon 9 and 8 with +9x and -8x so that(9x-8x) = x and (9x)X(-8x) = -72x^2
It is this finally -72x^2 =-(1X2X2X2X3X3)x^2 = (3X3x)X(-2X2X2x)]which you should try to get after assimilating all those small bits of things given above
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