Log base 2 (x-1) + log base 2 (x+9) = log base 2 (4x=3)

using log property

\( log_b~(a)+log_b~(c)\Leftrightarrow log_b~(a*c) \)

and

\( log_b~(a)-log_b~(c)\Leftrightarrow log_b~(\frac{a}{c}) \)

and

\( log_b~(a)=c\Leftrightarrow a=b^c \)

\( log_2~(x-1)+log_2~(x+9)=log_2~(4x+3) \)

\( log_2~[(x-1)*(x+9)]=log_2~(4x+3) \)

\( log_2~[(x-1)*(x+9)]-log_2~(4x+3)=0 \)

\( log_2~\frac{(x-1)*(x+9)}{(4x+3)}=0 \)

\( \frac{(x-1)*(x+9)}{(4x+3)}=2^0 \)

now

\( (x-1)*(x+9)=(4x+3) \)

\( x^2+8x-9=4x+3 \)

\( x^2+8x-9-4x-3=0 \)

\( x^2+4x-12=0 \)

\( \boxed{\boxed{X_1=-6~~and~~X_2=2}} \)

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