You need to take logarithms both sides such that:
`log(x^(2+log x)) = log1000`
You need to solve the following logarithmic identity such that:
`log_b(x^a) = a*log_b (x)`
Using this identity in case of the term `log(x^(2+log x))` yields:
`log(x^(2+log x)) = (2+log x)log x`
`(2+log x)log x = log 10^3`
You need to open the brackets to the left side such that:
`2log x + log^2 x = 3 log 10`
Since `log 10 = 1` , yields:
`log^2 x + 2log x - 3 = 0`
You should come up with the substitution `log x = y` such that:
`y^2 + 2y - 3 = 0`
You need to use quadratic formula such that:
`y_(1,2) = (-2+-sqrt(4 + 12))/2`
`y_(1,2) = (-2+-sqrt16)/2`
`y_(1,2) = (-2+-4)/2`
`y_1 = 1 ; y_2 = -6/2 =gt y_2 = -3`
You need to solve the equations `log x = y_1` and `log x = y_2` such that:
`log x = 1 =gt x = 10`
`log x = -3 =gt x = 10^(-3) =gt x = 1/(10^3) =gt x = 1/1000`
You should remember that logarithmic function is inverse to exponential function, hence, since the domain of exponential function is R, thus, the range of logarithmic function is R.
Hence, to avoid any confusion, you need to remember that the values of logarithmic function may be either positive, or negative, while, the values of x need to be positive.
To prove the statement above, you need to remember the equation of exponential function such that:
`a^x = b`
Notice that x may be positive or negative, while b is always positive, for any real x.
You need to take logarithms of both sides such that:
`log_a (a^x) = log_a b`
Using logarithmic identity yields:
`x log_a a = log_a b =gt x = log_a b`
Substituting `log_a b` for x in equation `a^x = b` yields:
`a^(log_a b) = b`
Notice that `log_a b` substitutes x, hence, `log_a b` may be either positive, or negative.
Since both values of x satisfy the condition of existence of logarithms, x>0, then you need to state that the equation has two solutions.
Hence, evaluating the solutions to the given equation yields `x = 1/1000` and `x = 10` .