Answer:
[tex]RHS=LHS[/tex]
Step-by-step explanation:
[tex]3^x=4^y[/tex]
Taking log both side,
[tex]log_{e}3^x=log_{e}4^y[/tex]
[tex]xlog_{e}3=ylog_{e}4[/tex]
[tex]\frac{x}{y}log_{e}3=log_{e}4[/tex]
[tex]\frac{x}{y}=\frac{log_{e}4}{log_{e}3}=\frac{x}{y}=log_{e}(4-3)=log_{e}(1)=0[/tex]
[tex]x=0[/tex],
therefore [tex]y=0, z=0[/tex]
To Prove
[tex]z(x+y)=xy[/tex]
on putting values of x, y and z in above equation,
[tex]0=0[/tex]
Hence proved
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Answers & Comments
Answer:
[tex]RHS=LHS[/tex]
Step-by-step explanation:
[tex]3^x=4^y[/tex]
Taking log both side,
[tex]log_{e}3^x=log_{e}4^y[/tex]
[tex]xlog_{e}3=ylog_{e}4[/tex]
[tex]\frac{x}{y}log_{e}3=log_{e}4[/tex]
[tex]\frac{x}{y}=\frac{log_{e}4}{log_{e}3}=\frac{x}{y}=log_{e}(4-3)=log_{e}(1)=0[/tex]
[tex]x=0[/tex],
therefore [tex]y=0, z=0[/tex]
To Prove
[tex]z(x+y)=xy[/tex]
on putting values of x, y and z in above equation,
[tex]0=0[/tex]
[tex]RHS=LHS[/tex]
Hence proved