Solve Logarithms: Find X When 8logx-3logx^2=log2

[donniemateno]

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

 

[HallsofIvy]

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

Surely this is supposed to be log(x^2), not (log x)^2 so this is NOT a quadratic equation.

log(x^2)= 2log(x) so you have 8log(x)- 6 log(x)= 2log(x)= log 2.

2lo(x)= log(x^2)= log(2) so x^2= 2.

= -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

 

[Mark44]

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

You're treating the equation below as a quadratic equation. It's not, so you can't use the Quadratic Formula.

= 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

Incidentally, don't connect equations with =. Instead, use "implies" - ==> or "is equivalent to" - <==>.
8logx-3logx² = log8x - log4x
==> 8logx-3logx² = log2

In the next step replace 8logx and 3logx² by different expressions using one of the properties of logarithms. In each case you should get log <something>, with no coefficient in front of log.

In the step after that combine log <A> - log<B> into a single log expression. A and B are the "somethings" you get in the previous step.

If all goes well, you should end up with an equation that looks like this:
log<something else> = log 2.

If log a = log b, what can you say about a and b?