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Q.
If $\log _{\left(5.2^{ x }+1\right)} 2 ; \log _{\left(2^{1-x}+1\right)} 4$ and 1 are in Harmonical Progression then
Sequences and Series
Solution:
$a , b , c$ are in H.P. $\Rightarrow b =\frac{2 ac }{ a + c } \Rightarrow \frac{\log 4}{\log \left(2^{1- x }+1\right)}=\frac{2 \cdot \frac{\log 2}{\log \left(5 \cdot 2^{ x }+1\right)} \cdot 1}{\frac{\log 2}{\log \left(5 \cdot 2^{ x }+1\right)}+1}$
$\frac{2 \log 2}{\log \left(2^{1-x}+1\right)}=\frac{2 \log 2}{\log \left(5 \cdot 2^{ x }+1\right)\left[\log 2+\log \left(5 \cdot 2^{ x }+1\right)\right.}$
$. t +2=2 / t +1 \Rightarrow 10 t ^2+2 t =2+ t\left(2^{ x }= t \right) $
$10 t ^2+ t -2=0$
$10 t ^2+5 t -4 t -2=0 $
$5 t (2 t -1)-2(2 t +1)=0 \Rightarrow t =2 / 5,-1 / 2$
$x \log 2=\log 2 / 5 \Rightarrow 2^{ x }=2 / 5$
$x \log _2 2=1-\log _2 5 $
$x =1-\log _2 5 .$