Question

how to do this problems​

Answer 1

Answer:

your answer is in attachment

Step-by-step explanation:

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Answer 2

$5.\\\\x^2+y^2=25xy\\\\\textsf {Adding 2xy on both sides,}\\\\x^2+y^2+2xy=25xy+2xy\\\\(x+y)^2=27xy\\\\\log[(x+y)^2]=\log(27xy)\\\\2\log(x+y)=\log (3^3\cdot x\cdot y)\\\\\boxed {2\log (x+y)=3\log 3+\log x+\log y}$

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$6.\\\\\log\left (\dfrac {x+y}{3}\right)=\dfrac {1}{2}(\log x+\log y)\\\\\\\log (x+y)-\log 3=\dfrac {1}{2}\log (xy)\\\\\\2[\log (x+y)-\log 3]=\log (xy)\\\\\log [(x+y)^2]-\log 9=\log (xy)\\\\\log [(x+y)^2]-\log (xy)=\log 9\\\\\\\log\left (\dfrac {(x+y)^2}{xy}\right)=\log 9\\\\\\\textsf {Taking antilogs,}\\\\\\\dfrac {x^2+y^2}{xy}+2=9\\\\\\\dfrac {x^2+y^2}{xy}=9-2\\\\\\\boxed {\dfrac {x}{y}+\dfrac {y}{x}=\mathbf {7}}$

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$7.\\\\(2.3)^x=1000\quad\implies\quad\log_{2.3}(1000)=x\\\\\\\dfrac {\log 1000}{\log\left (\dfrac {23}{10}\right)}=x\\\\\\\dfrac {3}{\log 23-1}=x\\\\\\\dfrac {\log 23-1}{3}=\dfrac {1}{x}\\\\\\\textsf{And,}\\\\(0.23)^y=1000\quad\implies\quad\log_{0.23}(1000)=y\\\\\\\dfrac {\log 1000}{\log\left (\dfrac {23}{100}\right)}=y\\\\\\\dfrac {3}{\log 23-2}=y\\\\\\\dfrac {\log 23-2}{3}=\dfrac {1}{y}$

$\textsf {Now,}\\\\\\\dfrac {1}{x}-\dfrac {1}{y}=\dfrac {\log 23-1}{3}-\dfrac {\log23-2}{3}\\\\\\\dfrac {1}{x}-\dfrac {1}{y}=\dfrac {\log 23-1-\log 23+2}{3}\\\\\\\boxed {\dfrac {1}{x}-\dfrac {1}{y}=\mathbf {\dfrac {1}{3}}}$