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We are given that $x^{\log x}=100x.$ If we take the base-10 logarithm of each side, we have

$\log\left(x^{\log x}\right) = \log(100x).$

Thus,

$\log x \cdot \log x = \log 100 + \log x,$

or

$\log x \cdot \log x = 2+\log x.$

If we let $n = \log x,$ then our equation becomes $n^2 = 2+n,$ which factors as $(n-2)(n+1) = 0.$ Thus, $n=2$ or $n=-1,$ giving $x = 100$ or $x=\frac{1}{10},$ respectively.

Thus, there are two solutions to the equation.

SOLUTION

Let us substitute $log_{10}x = k$ . ----- Equation (1)

We then also have $x = 10^{k}$ . ----- Equation (2)

Substituting Equations (1) and (2) in the given equation $x ^{\log_{10} x } = 100x$ , we get

$(10^{k})^{k} = 100(10^{k}) \Rightarrow 10^{k^{2}} = 10^{k+2}$

$\Rightarrow k^{2} = k+2 \Rightarrow (k+1)(k-2) = 0 \Rightarrow k = (-1)$ or $k = 2$ .

$k = 2 \Rightarrow x = 10^{2} = 100$ .

$k = (-1) \Rightarrow x = 10^{-1} = 0.1$ .

Hence, there are only $\boxed{2}$ solutions for the given equation.

Take log to the base 10 on both sides and then simplify by taking all terms on LHS. $\log _{ 10 }{ x } \times \quad \log { x } =\quad \log { 10 } +\quad \log { x } \\ \log _{ 10 }{ x } \times \quad \log { x } -\log { 10 } -\log { x } =\quad 0\\ \log { x } \left( \log _{ 10 }{ x\div 10 } \right) =0$

hence answers are 1 and 10

let log x = y,then x=10^y, so the eqn becomes now: (10^y)^y = 100 x 10^y =10^(y+2) so, 10^(y^2) = 10^(y+2) i.e y^2 - y - 2 = 0, solving this eqn we get, y=-1,2 ( 2 solutions)

Taking logarithms both the sides to the base 10 Take logx =t Solve the quadratic Two roots will be there.

We can start by taking log of both sides

Log (x^(log x)) = log (100x)

(Log x )(log x) = log 100 + log x (using properties of logs)

(Log x)^2 = 2 + log x

Making a quadratic equation with logs, we have

(Log x)^2 - log x - 2 = 0

This equation factors to

(Log x - 2)(log x + 1) = 0

Log x = 2 , log x = -1, both can exist in real numbers, hence 2 solutions