Math, asked by venkataharshitha, 1 month ago

6. If log(x+y÷3)=1÷2(logx+logy),then find the value of x÷y+y÷x?​

Answers

Answered by ItzAshi
157

Step-by-step explanation:

{\large{\mathbf{\underline{\blue{Question :→}}}}} \\

If log (x + y)/3 = 1/2(log x + log y) then find the value of x/y + y/x

{\large{\mathbf{\underline{\blue{Solution :→}}}}} \\

We have,

 {\bold{\sf{log \:  \frac{x \:  + \:  y}{3} \:  = \:  \frac{1}{2}(log  x  \: +  \: log y)}}} \\  \\

 {\bold{\sf{: \:  ⟹ \:  \:  \:  \:  \:  log \frac{x \:  + \:  y}{3}  \:  \: = \:  \:  (log x) ^ \frac{1}{2}  \: +  \: (log y) ^ \frac{1}{2}}}} \\  \\

 {\bold{\sf{: \:  ⟹ \:  \:  \:  \:  \:  log \frac{x  \: +  \: y}{3}  \:  \: = \:  \:  log √x  \: + \:  log √y}}} \\  \\

{\bold{\sf{:  \: ⟹ \:  \:  \:  \:  \:  log \frac{x  \: + \:  y}{3}  \:  \: = \:  \:  log  \sqrt{x}  \:  \sqrt{y} }}} \\  \\

{\bold{\sf{:  \: ⟹  \:  \:  \:  \:  \: \frac{x \:  + \:  y}{3}  \:  \: = \:  \:  \sqrt{xy}}}} \\  \\

{\bold{\sf{: \:  ⟹ \:  \:  \:  \:  \:  x  \: +  \: y  \:  \: =  \:  \: 3 \: \sqrt{xy}}}} \\  \\

On squaring both the sides we will get,

{\bold{\sf{:  \: ⟹  \:  \:  \:  \:  \: (x  \: + \:  y)²  \:  \: = \:  \:  9xy}}} \\  \\

{\bold{\sf{:  \: ⟹  \:  \:  \:  \:  \:  x²  \: + \:  y²  \: +  \: 2xy  \:  \: = \:  \:  9xy}}} \\  \\

{\bold{\sf{: \:  ⟹  \:  \:  \:  \:  \: x²  \: +  \: y² \:  =  \: 9xy  \: -  \: 2xy}}} \\  \\

{\bold{\sf{\red{:  \: ⟹ \:  \:  \:  \:  \: }}}}{\bold{\mathfrak{\fbox{\orange{ x²  \: +  \: y²  \: = \:  7xy}}}}} \\  \\

On dividing both sides by xy we will get,

{\bold{\sf{:  \: ⟹ \:  \:  \:  \:  \:  \frac{x²  \: + \:  y²}{xy}  \:  \: =  \:  \: \frac{7xy}{xy}}}} \\  \\

{\bold{\sf{\red{: \:  ⟹ \:  \:  \:  \:  \: }}}}{\bold{\mathfrak{\boxed{\purple{ \frac{x}{y}  \: +  \: \frac{y}{x}  \:  \: =  \:  \: 7}}}}} \\

Answered by muskansingh3707126
2

Step-by-step explanation:

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Step-by-step explanation:

{\large{\mathbf{\underline{\blue{Question :→}}}}} \\

If log (x + y)/3 = 1/2(log x + log y) then find the value of x/y + y/x

{\large{\mathbf{\underline{\blue{Solution :→}}}}} \\

We have,

 {\bold{\sf{log \:  \frac{x \:  + \:  y}{3} \:  = \:  \frac{1}{2}(log  x  \: +  \: log y)}}} \\  \\

 {\bold{\sf{: \:  ⟹ \:  \:  \:  \:  \:  log \frac{x \:  + \:  y}{3}  \:  \: = \:  \:  (log x) ^ \frac{1}{2}  \: +  \: (log y) ^ \frac{1}{2}}}} \\  \\

 {\bold{\sf{: \:  ⟹ \:  \:  \:  \:  \:  log \frac{x  \: +  \: y}{3}  \:  \: = \:  \:  log √x  \: + \:  log √y}}} \\  \\

{\bold{\sf{:  \: ⟹ \:  \:  \:  \:  \:  log \frac{x  \: + \:  y}{3}  \:  \: = \:  \:  log  \sqrt{x}  \:  \sqrt{y} }}} \\  \\

{\bold{\sf{:  \: ⟹  \:  \:  \:  \:  \: \frac{x \:  + \:  y}{3}  \:  \: = \:  \:  \sqrt{xy}}}} \\  \\

{\bold{\sf{: \:  ⟹ \:  \:  \:  \:  \:  x  \: +  \: y  \:  \: =  \:  \: 3 \: \sqrt{xy}}}} \\  \\

On squaring both the sides we will get,

{\bold{\sf{:  \: ⟹  \:  \:  \:  \:  \: (x  \: + \:  y)²  \:  \: = \:  \:  9xy}}} \\  \\

{\bold{\sf{:  \: ⟹  \:  \:  \:  \:  \:  x²  \: + \:  y²  \: +  \: 2xy  \:  \: = \:  \:  9xy}}} \\  \\

{\bold{\sf{: \:  ⟹  \:  \:  \:  \:  \: x²  \: +  \: y² \:  =  \: 9xy  \: -  \: 2xy}}} \\  \\

{\bold{\sf{\red{:  \: ⟹ \:  \:  \:  \:  \: }}}}{\bold{\mathfrak{\fbox{\orange{ x²  \: +  \: y²  \: = \:  7xy}}}}} \\  \\

On dividing both sides by xy we will get,

{\bold{\sf{:  \: ⟹ \:  \:  \:  \:  \:  \frac{x²  \: + \:  y²}{xy}  \:  \: =  \:  \: \frac{7xy}{xy}}}} \\  \\

{\bold{\sf{\red{: \:  ⟹ \:  \:  \:  \:  \: }}}}{\bold{\mathfrak{\boxed{\purple{ \frac{x}{y}  \: +  \: \frac{y}{x}  \:  \: =  \:  \: 7}}}}} \\

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