ElectricPavlov

sqrt(2x-3) = x-3

2x-3 = (x-3)^2

2x-3 = x^2-6x+9

x^2-8x+12 = 0

(x-6)(x-2) = 0     so x = 2 or 6      Do both answers solve the equation?  Or is one of them extraneous?   What do YOU think?

Using the log base change rule to change to log_{10 :}

log x / log3  = (-2 + log100/log2)(log(sqrt2)/log3)

Now, using calculator you will find x = 5

y = 2 + 20/6 x - 3y    add 3 y to both sides of the equation

4y = 2 + (20/6) x        your original Q says y = 3    substitute that in

12 = 2 + 20/6 x       subtract 2 from both sides

10 = 20/6   x            multiply both sides by 6/20

60/20 = x = 3

Now I see your question has TWO Equations....please put some spaces in there !

y = 2 + 5/6 * x

4x - 3y = 3          Is much more clear.....see whoisjoe answer below....

f/(r ^ 2)  = n + k - a     after subtracting  'a' from both sides of the equation

f/(n+k-a) = r^2      after re-arranging    now take sqrt of both sides

r = +- sqrt ( f/(n+k-a) )       (and as WIJ pointed out   n+k-a cannot equal 0)

  If you can find very dark skies with no moon, you can see an ARM of the spiral Milky Way stretching from north to south across the sky....looks kinda like a thin high sirrus cloud....here is a couple of pics from last Summer: (one even has a faint meteor streaking across it)

log₁₀ 10000 = y

From the definition of logs:

Log₁₀  x = y         exactly if 10^y    =   x

log₁₀ 10000 = y        if    10^y =10000         for 10^y=10000    ,    y MUST = 4

Another method:

2^(2x+1) =2^2x *2¹

Sub in to the original equation

2^2x * 2   +  2^2x  = 3/8

2^2x * 3 = 3/8

2^2x = 1/8       take log of both sides

2x log 2 = log 1/8

x =  1/2 *  log (1/8) / log2   = -1.5

We know angle of incidence = angle of reflection, but we'll ned the diagram of the room to find the third wall strike point !  

Hey Melody!  

  That is what I did (on paper) to find the other root.....I got the same numbers as you did for a and b.  I just didn't feel like transcribing it all on the keyboard when I was done !    (Maybe for a REGISTERED user who wanted derivation I would have...)

 The question states a and b are real.....well , yes they are, but 'b' is part of   b i   (which obviously is not real)...a bit misleading, but at least we both figured it out....