Answer:
or
Explanation:
We are given a logarithmic equation of x and we have to solve it for x.
Given,
![x^(\log x) = 1000000x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/e04up0vrxwv7hkgn786clpf59k5holtvsu.png)
⇒
![x^(\log x) = 10^(6) * x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pba0un45qon1ik4dia07n547vc0e7pco8f.png)
Now, taking log on both sides, we get
![\log x^(\log x) = \log 10^(6) * x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pmsleg40n0sqbg2tujfto1rzjrb5hzizyo.png)
⇒
![\log x .\log x = \log 10^(6) + \log x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hrkwx7gxdn075erifykc14texz8g3qh5zy.png)
{Since
and
}
⇒
![(\log x)^(2) = 6 + \log x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/see9uxu8cwkuwavtj8xqeumi138hjmhdkg.png)
{Since log 10 = 1}
⇒ a² = 6 + a {Where, a = log x}
⇒ a² - a - 6 = 0
⇒ (a - 3)(a + 2) = 0
⇒ a = 3 or a = -2
⇒
or
![\log x = - 2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/u5r4c8ln48on9b05upkgrwuo3sqp42eshe.png)
Now, converting logarithm to exponential form, we get,
⇒
or
(Answer)