Log_5(x+12)>=log_5(4x+3)
=log_5(4x+3)\\
x+12>=4x+3\\
x+12>0\\
4x+3>0\\
3x<=9\\x>-12\\
4x>-3\\
x<=3\\x>-12\\x>-3/4\\
(-3/4;3]\\C" alt="log_5(x+12)>=log_5(4x+3)\\
x+12>=4x+3\\
x+12>0\\
4x+3>0\\
3x<=9\\x>-12\\
4x>-3\\
x<=3\\x>-12\\x>-3/4\\
(-3/4;3]\\C" align="absmiddle" class="latex-formula">