- #1
donniemateno
- 45
- 0
logarithms STUCK!
8logx-3logx^2 = log8x - log4x
im struggling to find x. my working out so far is:
rhs log8x-log4x
= log2
= 8logx - 3logx^2 = log 2
= 8logx - 3logx^2 - log 2 = 0
= -b+ or - the square root of b^2-4ac divided by 2a
a = -3 b = 8 c = - 2
= -8 + or - sqaure root of 8^2 (-4x-3x^2-2) divided by 2x - 3^2
x= -8 + or - square root of 64 - 24 divided by -6
= square of 40 divided by -6
= -8 + or - 6.3245 divided by -6
= - 8 + or - , -1.05408
x= -9.0540
x= 6.94592
can anyone help me. am stuck and can't see where I've gone wrong
8logx-3logx^2 = log8x - log4x
im struggling to find x. my working out so far is:
rhs log8x-log4x
= log2
= 8logx - 3logx^2 = log 2
= 8logx - 3logx^2 - log 2 = 0
= -b+ or - the square root of b^2-4ac divided by 2a
a = -3 b = 8 c = - 2
= -8 + or - sqaure root of 8^2 (-4x-3x^2-2) divided by 2x - 3^2
x= -8 + or - square root of 64 - 24 divided by -6
= square of 40 divided by -6
= -8 + or - 6.3245 divided by -6
= - 8 + or - , -1.05408
x= -9.0540
x= 6.94592
can anyone help me. am stuck and can't see where I've gone wrong