# number of real solution of logarithmic and exponential equation

Number of real solution of $$\log_{\frac{1}{16}}(x) = \left(\frac{1}{16}\right)^x$$

$\bf{My\; Try::}$ Using hit and trial Here $\displaystyle x= \frac{1}{4}$ and $\displaystyle x = \frac{1}{2}$ are the solution of above equation.

Now for the existance of other solution, Let $$f(x) = \log_{\frac{1}{16}}(x) - \left(\frac{1}{16}\right)^x$$

So $$f'(x)=\log_{\frac{1}{16}}(e)\cdot \frac{1}{x}-\left(\frac{1}{16}\right)^x\cdot \log_{e}\left(\frac{1}{16}\right)$$

And $$f''(x)=-\log_{\frac{1}{16}}(e)\cdot \frac{1}{x^2}-\left(\frac{1}{16}\right)^x\cdot \log^2_{e}\left(\frac{1}{16}\right)$$

Now for $\max$ or $\min,$ Put $$\displaystyle f'(x)=0\Rightarrow \log^2_{\frac{1}{16}}(e) = x\cdot \left(\frac{1}{16}\right)^x$$

Now how can i calculate after that, Help required, Thanks

$f'(x)=0$ has explicit solutions in terms of Lambert function. The Wikipedia page shows many examples of the manipulations to be done.
In your case, the solutions are given by $$x_1=-\frac{W\left(-\frac{1}{\log (16)}\right)}{\log (16)}\approx 0.293610$$ $$x_2=-\frac{W_{-1}\left(-\frac{1}{\log (16)}\right)}{\log (16)}\approx 0.437246$$