Question

1 If x² + y2 = 7xy, then prove that log( x + y)/ 3 = 1/2 (log x +logy). ​

Answer 1

Given :

• $\sf\:x^{2}+y^{2}=7xy$

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To do :

• We need to prove $\sf\:\frac{log(x+y)}{3}=\frac{1}{2}(logx+logy)$

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Things to know ‎:

As in the given question there is log we need to know some basic properties of log. Let's have a look :

$\tt\:\small[1\small]$ $\sf\:log(xy) =logx+logy$

$\tt\:\small[2\small]$ $\sf\:log(\frac{x}{y}) =logx-logy$

$\tt\:\small[3\small]$ $\sf\:log(x^{n}) =nlogx$

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Solution :

$\sf\:x^{2}+y^{2}=7xy$

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Adding 2xy on both sides of the equation

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$\sf\::\implies\:x^{2}+y^{2}+2xy=7xy+2xy$

$\sf\::\implies\:x^{2}+y^{2}+2xy=9xy$

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Closely observing we get to know that LHS denotes an algebraic expression that has got a formula as

⠀⠀⠀⠀⠀⠀⠀⠀$\sf\:(a+b)^{2}=a^{2}+b^{2}+2ab$

Only difference here is x in place of a and y in place of b.

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$\sf\::\implies\:(x+y)^{2}=9xy$

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Taking log on both sides

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$\sf\::\implies\:log(x+y)^{2}=log (9xy)$

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Using third property of log in LHS as stated before

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$\sf\::\implies\:2log(x+y)=log (9xy)$

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9 in RHS can be written as 3²

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$\sf\::\implies\:2log(x+y)=log (3^{2}xy)$

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Using first property of log in RHS as stated before

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$\sf\::\implies\:2log(x+y)=log 3^{2} + log x + log y$

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Using third property of log in RHS as stated before

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$\sf\::\implies\:2log(x+y)=2log 3 + log x + log y$

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Dividing both sides by 2

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$\sf\::\implies\:\frac{\cancel{2}log(x+y)}{\cancel{2}}=\frac{2log 3 + log x + log y}{2}$

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$\sf\::\implies\:log(x+y)=\frac{2log 3 + log x + log y}{2}$

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$\sf\::\implies\:log(x+y)-2log3=\frac{log x + log y}{2}$

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Using second property of log in LHS as stated before

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$\sf\::\implies\:log\frac{(x+y)}{3}=\frac{log x + log y}{2}$

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

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‎⠀⠀⠀⠀$\bf\:Hence~Proved!~$

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Note :

⠀⠀⠀⠀Looks complicated as I have explained each step, but it's very simple‎. You can skip some of the basic steps.

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Answer 2

Answer:

You can’t, because it isn’t true!

You have two errors in your question.

First, x+y/3=x+y3

The correct formulation is: (x+y)/3=x+y/3

This comes from the convention on the order of mathematical operations, something you should have learnt before you were 10.

Second, remove either the ‘1/2’ or the ‘^2’ - one of them shouldn’t be there. I’m going to remove the former, so the correct equation to prove is:(x+y)/3=x+y/3

log(x)+log(y)=log[(x+y/3)2]

Now, (x+y)2=x2+2xy+y2=(x2+y2)+2xy [Eq. 1]

From the question, x2+y2=7xy [Eq. 2]

From Eq. 1 & 2, we thus have: (x+y)2=7xy+2xy=9xy

Dividing throughout by 9:

xy=(x+y)29=(x+y)232=(x+y/3)2

Taking logs of both sides (it doesn’t matter what base of logs we use, so I’ll use the generic ‘log’)

log(xy)=log[(x+y/3)2]

⇒log(x)+log(y)=log[(x+y/3)2]