[tex]log_512=a[/tex]
[tex]log_52=b[/tex]
a)
[tex]log_56=log_5\frac{12}{2}=log_512-log_52=a-b[/tex]
b)
[tex]log_59=log_53^2=2log_53=2log_5\frac{12}{4}=2(log_512-log_54)=[/tex]
[tex]2(a-log_52^2)=2(a-2log_52)=2(a-2b)[/tex]
c)
[tex]log_5\sqrt[5]3=log_53^{\frac{1}{5}}=\frac{1}{5}log_53=\frac{1}{5}log_5\frac{12}{4}=[/tex]
[tex]\frac{1}{5}(log_512-log_54)=\frac{1}{5}(a-log_52^2)=\frac{1}{5}(a-2log_52)=\frac{1}{5}(a-2b)[/tex]
d)
[tex]log_350=\frac{log_550}{log_53}=\frac{log_5(5\cdot5\cdot5)}{log_5(12:4)}=\frac{log_55+log_55+log_52}{log_512-log_54}=\\\\\frac{1+1+b}{a-log_52^2}=\frac{2+b}{a-2log_52}=\frac{2+b}{a-2b}[/tex]
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Verified answer
[tex]log_512=a[/tex]
[tex]log_52=b[/tex]
a)
[tex]log_56=log_5\frac{12}{2}=log_512-log_52=a-b[/tex]
b)
[tex]log_59=log_53^2=2log_53=2log_5\frac{12}{4}=2(log_512-log_54)=[/tex]
[tex]2(a-log_52^2)=2(a-2log_52)=2(a-2b)[/tex]
c)
[tex]log_5\sqrt[5]3=log_53^{\frac{1}{5}}=\frac{1}{5}log_53=\frac{1}{5}log_5\frac{12}{4}=[/tex]
[tex]\frac{1}{5}(log_512-log_54)=\frac{1}{5}(a-log_52^2)=\frac{1}{5}(a-2log_52)=\frac{1}{5}(a-2b)[/tex]
d)
[tex]log_350=\frac{log_550}{log_53}=\frac{log_5(5\cdot5\cdot5)}{log_5(12:4)}=\frac{log_55+log_55+log_52}{log_512-log_54}=\\\\\frac{1+1+b}{a-log_52^2}=\frac{2+b}{a-2log_52}=\frac{2+b}{a-2b}[/tex]