solve the equation?

Question

log₂(6-x) = 3 - log₂x

thanks

Answer 1

Is that a log with base of 2 or log2?

If it is log with base of 2. then;

In order to solve this equation for x, you must take the anti log of both sides in order to take x out of the log. The anti log of log₂ is 2^, so therefore:

2^log₂(6-x) = 2^3 - 2^log₂x

6-x=8-x

6=8

This equation have no right solution for x.

If it is log of 2, then;
6log(2)-log(2)x=3-log(2)x

6log(2) - 3 = 0

The equation still have no solution.

Did you prehaps copy the equation wrong?

Answer 2

In solution, all logs are taken as log base 2
log (6 - x) + log x = 3
log [ (6 - x).x ] = 3
6x - x² = 8 (where 8 = 2³)
x² - 6x + 8 = 0
(x - 4).(x - 2) = 0
x = 2, x = 4

Answer 3

I cant see the base. I'll assume it is 'a' for argument's sake AND that the logs are the same.

log(base a)(6-x) = 3 - log(base a)(x)
make everything an exponent of a
a^(log(base a)(6-x)) = a^(3 - log(base a)(x))
(6-x) = a^3 / a^(log(base a) (x))
6-x= a^3/x
6x - x^2 = a^3
x^2 - 6x = -a^3

Answer: x^2 - 6x + a^3 = 0 Sustitute the a for the base of the log. Please see example for specific example.

Lets Assume a = 2
x^2 - 6x + 2^3 = 0
x^2 - 6x + 8 = 0
(x - 4) (x - 2)
x = 4 & 2

Answer 4

log₂(6-x) = 3 - log₂x
log₂(6-x) + log₂x = 3
log₂[(6-x)x] = 3
[(6-x)x] = 2^3
6x - x^2 = 8
0 = x^2 - 6x + 8
(x-2)(x-4) = 0
x = 2 or 4

Answer 5

log(base2)(6-x)=3-log(base2)x

we use: 3 = 3log(base2)(2)= log(base2)(2^3) = log(base2)(8)
<=> log(base2)(6-x)=log(base2)8 - log(base2)x

we use loga-logb=log(a/b)
<=> log(base2)(6-x)=log(base2)(8/x)

we use loga=logb <=> a=b
<=> 6-x=8/x

we multiply by x on both side (we rule out 0 as solution)
<=> -x^2+6x=8 (and x<>0)

we arrange the term of the equation to get something of the form (ax^2+bx+c=0)
<=> x^2-6x+8=0 (and x<>0) (*)

We resolve (*)...
delta=36-32=4 (=b^2-4ac)
2 solutions to the equation (*):
x1= (-b+sqrt(delta))/2a = (6+2)/2 = 4
x2= (-b-sqrt(delta))/2a = (6-2)/2 = 2

Possible solutions are 2 and 4

Answer 6

(6-x)log2 = 3-xlog2
6log2-xlog2 = 3-xlog2
work out what log2 is in your calculator and then sub in - i hope this works! good luck.