Question

title="5^{1+log_4x}+5^{log_{0.25}x-1}=\frac{26}{5}" alt="5^{1+log_4x}+5^{log_{0.25}x-1}=\frac{26}{5}" align="absmiddle" class="latex-formula">

Answer 1

$\displaystyle 5^{1+\log_4 x}+5^{-\log_4x-1} = \frac{26}{5}\\ \\ \text{Sea } z=5^{1+\log_4 x} \text{ entonces}: \\ \\ z+\frac{1}{z}=\frac{26}{5}\\ \\ 5z^2-26z+5=0 \\ \\ (5z-1)(z-5)=0 \\ \\ z\in\left\{\frac{1}{5};5\right\}\\ \\ \text{Then: }\\ \\ 5^{1+\log_4x}=5^{-1} \vee 5^{1+\log_4x}=5\\ \\ 1+\log_4x=-1 \vee 1+\log_4x=1\\ \\ \log_4x =-2 \vee \log_4x=0 \\ \\ x=\frac{1}{16} \vee x=1\\ \\ \text{Answer: }x\in\left\{\frac{1}{16};1\right\}$