Mathematics
shyqqurrl
6

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

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(2) Answers
katiebabyyy

using log property [latex]log_b~(a)+log_b~(c)\Leftrightarrow log_b~(a*c)[/latex] and [latex]log_b~(a)-log_b~(c)\Leftrightarrow log_b~(\frac{a}{c})[/latex] and [latex]log_b~(a)=c\Leftrightarrow a=b^c[/latex] [latex]log_2~(x-1)+log_2~(x+9)=log_2~(4x+3)[/latex] [latex]log_2~[(x-1)*(x+9)]=log_2~(4x+3)[/latex] [latex]log_2~[(x-1)*(x+9)]-log_2~(4x+3)=0[/latex] [latex]log_2~\frac{(x-1)*(x+9)}{(4x+3)}=0[/latex] [latex]\frac{(x-1)*(x+9)}{(4x+3)}=2^0[/latex] now [latex](x-1)*(x+9)=(4x+3)[/latex] [latex]x^2+8x-9=4x+3[/latex] [latex]x^2+8x-9-4x-3=0[/latex] [latex]x^2+4x-12=0[/latex] [latex]\boxed{\boxed{X_1=-6~~and~~X_2=2}}[/latex] I hope you enjoy it...

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