Wiedząc, że, jeżeli log x = log a, to x = a, oblicz x, gdy: a) log x = log4 + log5 + log6, b) log x = log40 - log5,  c) log x = 2log3 + log4,  d) log x= 1/2log5 + 1/2log3. Z góry dziękuję. ; ) 

a) [latex]log x = log4 + log5 + log6[/latex] [latex]log x = log(4cdot 5cdot 6)[/latex] [latex]log x = log 120[/latex] [latex]x=120[/latex] b) [latex]log x = log40 - log5[/latex] [latex]log x =log frac{40}{5}[/latex] [latex]log x =log 8[/latex] [latex]x=8[/latex]   c) [latex]log x = 2log3 + log4[/latex] [latex]log x = log3^2 + log4[/latex] [latex]log x = log9 + log4[/latex] [latex]log x = log(9cdot 4)[/latex] [latex]log x = log 36[/latex] [latex]x = 36[/latex]   d) [latex]log x=frac{1}{2}log5 + frac{1}{2}log3[/latex] [latex]log x=frac{1}{2}(log5 +log3)[/latex] [latex]log x=frac{1}{2}log(5cdot 3)[/latex] [latex]log x=frac{1}{2}log15[/latex] [latex]log x=log15^{frac{1}{2}}[/latex] [latex]x=15^{frac{1}{2}}[/latex] [latex]x= sqrt{15}[/latex]

[latex]a) logx=log4+log5+log6=log(4*5*6)=log120, x=120 \ b) logx=log40-log5=logfrac{40}{5}=log8, x=8 \ c) logx=2log3+log4=log3^2+log4=log(9*4)=log36, x=36 \ d) logx=frac{1}{2}log5+frac{1}{2}log3=logsqrt{5}+logsqrt{3}=logsqrt{15}, x=sqrt{15}[/latex]