Question

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

Answer 1

Answer in an annex
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Answer 2

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

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

$x^2+8x-9=4x+3$

$x^2+8x-9-4x-3=0$

$x^2+4x-12=0$

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

I hope you enjoy it...