A maths sec 3
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hi
can any1 tell mi wads teh answer of 3 log a (2) - 4 + log a (a to the power of 3) whe nconverted into a single logarithmic form
and value of x when 4 (x to the power of half) = 3-7x -
3 log(a,2) - 4 + log(a,a³)
= log(a,8) - log(a,a^4) + log(a,a³)
= log[a,(8/a)]
4*sqrt(x)=3-7x
Let z be sqrt(x), z>=0
7z²+4z-3=0
(7z-3)(z+1)=0
z = 3/7 or -1 (rej. as z>=0)
x = sqrt(3/7) -
how come liddat ar? 7z²+4z-3=0Originally posted by ^tamago^:3 log(a,2) - 4 + log(a,a³)
= log(a,8) - log(a,a^4) + log(a,a³)
= log[a,(8/a)]
4*sqrt(x)=3-7x
Let z be sqrt(x), z>=0
7z²+4z-3=0
(7z-3)(z+1)=0
z = 3/7 or -1 (rej. as z>=0)
x = sqrt(3/7) -
gt careless mistake... shld be
z = 3/7
x = (3/7)^2 = 9/49 -
sqrt(x) = zOriginally posted by gobez14:how come liddat ar? 7z²+4z-3=0
x = z²
4*sqrt(x)=3-7x
4z=3-7z²
7z²+4z-3=0
(7z-3)(z+1)=0
z = 3/7 or -1 -
oh yeah. fug~!Originally posted by SBS261P:gt careless mistake... shld be
z = 3/7
x = (3/7)^2 = 9/49 -
can elaborate?Originally posted by SBS261P:gt careless mistake... shld be
z = 3/7
x = (3/7)^2 = 9/49 -
4x^½ = 3 - 7xOriginally posted by gobez14:can elaborate?
Let x^½ be u
4u = 3 - 7u²
7u² + 4u - 3 = 0
(7u - 3)(u + 1) = 0
7u - 3 = 0 or u + 1 = 0
u = 3/7 or u = -1
x^½ = 3/7 or x^½ = -1
x = 9/49 or 1
FYI, x^½ is equals to "square root of x", written as "sqrt x" -
but hor y i sub 1 into the original equation 1 side is 4 the other side is -4 ?Originally posted by wishboy:4x^½ = 3 - 7x
Let x^½ be u
4u = 3 - 7u²
7u² + 4u - 3 = 0
(7u - 3)(u + 1) = 0
7u - 3 = 0 or u + 1 = 0
u = 3/7 or u = -1
x^½ = 3/7 or x^½ = -1
x = 9/49 or 1
FYI, x^½ is equals to "square root of x", written as "sqrt x"
liddat still can?? -
x cannot be 1 becoz
x^1/2 = -1 means
square root x = -1
but square root cannot give negative number so must reject