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# Exponential Equation

- Question 4
- Solving the equation x
^{lg x}= 1000 x^{2}

- Solution:
- From the definition of logarithm, if a
^{x}= b (a > 0, and a is not 0) then x = log_{a}b - If a = 10, then log
_{10}b, marked as lg b, - Take the lg of both side ( note: lg b is log
_{10}b) - lg(x
^{lg x}) = lg(1000 x^{2}) - (lg x)(lg x) = lg1000 + lg x
^{2} - (lg x)
^{2}= lg 10^{3}+ 2 lg x - (lg x)
^{2}- 2 lg x - 3 = 0 - Factoring this quadratic equation, (lg x + 1)(lg x - 3) = 0
- lg x + 1 = 0 or lg x - 3 = 0
- For lg x + 1 = 0 => lg x = -1 => 10
^{-1}= x => x = 1/10 - For lg x - 3 = 0 => lg x = 3 => 10
^{3}= x => x = 1000

- formulas used:
- log
_{a}b^{x}= x log_{a}b - log
_{a}a = 1 - log
_{a}(MN) = log_{a}M + log_{a}N