solve the following equation for the value x:?

I think you mean (logx)^3=log (x^3)

log(x^3)=3log x

log x=a then

a^3=3a

a^3-3a=0

a(a^2-3)=0

If a=0 then log x=0 and x=1

If a^2-3=0 then a^2=3 and a= + or - (3sqert)

If a=(3sqrt) then log x=(3 sqrt) and x=(base)^(3sqrt)

If a=-(3sqrt) then logx=-(3sqrt) and x=(base)^(-3 sqrt)

log x^3 = log x^3 is like saying 4 = 4. When both sides of the equation are identical, the equation is always true. It won't matter what you substitute for x.

So the answer is "all real solutions" or "all solutions" depending on the book or teacher.

Your notation is slightly confusing, but I'm going to assume that it is equivalent to log(x^3) = (log x)^3

In this case, using a logarithmic identity gives us log(x^3) = 3*log(x)

thus, we get 3(log x) = (log x)^3

dividing by log x, we get 3 = (log x)^2

take the square root to get log x = sqrt(3)

from there, x = 10^(sqrt(3)), which is approximately 10^1.73205081 or 53.9573746

(log x^3) = log x^3

The solution is for all x > 0.

If you mean (log x)^3 = log x^3, then

(log x)^3 - 3log x = 0, x > 0

log x [(log x)^2 - 3] = 0

log x = 0, => x = 1

log x = ±√3, => x = 10^[±√3]

(log x^3) = log x^3

log x^3 = log x^3

but x cannot be 0 or negative since logx where x<= 0 is undefined thus,

x>0

5 - 2(x-a million) = 3x - 8 => 5 + 8 = 3x + 2(x-a million) (positioned all the X's on one area) => 5 + 8 = 3x + 2x - 2 (multiply out the brackets) => 5 + 8 + 2 = 3x + 2x (pass the two over to the different area) => 15 = 5x (upload each little thing on the two facets) => x = 15/5 (divide) => x = 3