CPhill

2 log₂ x - log₂ (x-1/2) = -log_1/3 3       note  that     - log_1/3 3     actually evaluates to  -[ -1 ]   = 1

So we have

2 log₂ x - log₂ (x-1/2)   = 1       and we can write

log₂ x²  - log₂ ( x - 1/2)   = 1       and by a property of logs  we have

log₂  [ ( x^2 ) / (x - 1/2)]    =  1    and in exponential form we have

2^1  =  ( x^2 ) / (x - 1/2)

2  = ( x^2) / ( x - 1/2)         multiply both sides by  x - 1/2

2( x - 1/2)  = x^2

2x - 1   = x^2       simplify

x^2 - 2x + 1  =  0        factor

(x -1 ) ( x - 1)  = 0      and setting each factor to 0   .....we have that   x = 1

Check that

2 log₂ (1) - log₂ (1-1/2)   = 1 

2 * 0   -  log₂ (1/2)  = 1

0  -  (-1)   = 1

1  =   1

Very nice, Riddler and Anonymous4338....5 pts.....!!!!

Thanks, Melody for that answer....

Here's another way to evalute this for    i    raised to any  "n"   positve integer

n mod 4  =  1    →   i

n mod 4  =  2   →   -1

n mod 4  = 3   →   - i

n mod 4  = 4   →    1

So

i^61  =      61 mod 4   = 1    →   i            just as Melody found   !!!

I thimk you have this

s / 6   =  209 / s         and both sides   = 11

So....

s / 6   = 11    multiply both sides by 6      and s = 66 on the left side

209 / s  = 11    we can rearrange as      209 / 11   =   s   = 19      and this is the value of s on the right ....check for yourself that :

66 / 6   = 209  / 19       =   11

13  is the coefficient

"a"  is the variable

cos (1/6)pi   = cos 30°  =  sqrt (3) / 2

4.

-(x-2)=x+2   simplify

-x + 2  = x + 2       subtract 2 from each side

-x  =  x          add x to both sides

0 = 2x           divide both sides by 2

0 = x

3.

2+3x=x+2+2x        simplify

2 + 3x   =  3x + 2        these two sides are exactly the same.......when we have this situation ........we have infinite solutions for "x"

I think there's a mistype in 1

The  objective in these problems is to get a variable expression [ a number with an x attached to it ] on one side of the equation  and a constant  [ a number  without the x]  on the other side

Here's 2

 -2x - 5 =   4x + 2       add 2x to both sides.....subtract 2 from  both sides

+2x - 2 = +2x -  2

       -7  =  6x             divide  each side by 6

     -7/6  = x

Hit the "2nd"   key.....you will see.....asin   acos   atan etc......these are the inverse trig functions