Respuesta :

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.