Answer:
x = ±2
Step-by-step explanation:
A equation is given to us , which is ,
[tex]\longrightarrow 2log_5(x) = log_5 4 [/tex]
From properties of logarithm we know that ,
[tex]\longrightarrow alog\ m = log \ m^a [/tex]
Applying this to LHS , we have ;
[tex]\longrightarrow log_5 x^2 = log_5 4 [/tex]
Now the bases of logarithm on LHS and RHS is same . On comparing , we have ;
[tex]\longrightarrow x^2 = 4 [/tex]
Put square root on both sides,
[tex]\longrightarrow x =\sqrt{4}[/tex]
Simplify ,
[tex]\longrightarrow \underline{\underline{ x =\pm 2 }}[/tex]
This is the required answer.