Question

if log(x^2 y^2)=9,log(x/y)=2 show that log x=3 log y=1​

Answer 1

Step-by-step explanation:

log (xy)² = 8

2 (log x + log y) = 8

log x+ log y = 4 --A

log x- log y =2 --B

A+B

2 logx = 6

log x= 6/2 = 3

A-B

2logy = 2

log y = 1

for log x =3, log y= 1 then log (x²y²) should be = 8

Answer 2

Correct question :

If log(x² y²) = 8 , log(x/y)=2 show that log x=3 log y=1

Given :

• log(x²y²) = 8
• log(x/y) = 2

To prove :

• log(x) = 3
• log(y) = 1

Formula used :

$\sf \bullet \: \: log( {a}^{m} ) = \: m\times log(a ) \\ \\ \sf \bullet \: \: log( \dfrac{a}{b} ) \: = \: log(a) - log(b) \\ \\ \sf \bullet \: \: log(ab) = log(a) + log(b)$

Solution :

• log(x²y²) = 8

➝ log(x²) + log(y²) = 8

➝ 2 log(x) + 2 log(y) = 8

➝ 2{ log(x) + log(y) } =

➝ log(x) + log(y) = 8/2

➝ log(x) + log(y) = 4 ...... equation 1

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$\sf \bullet \: \: log( \dfrac{x}{y} ) \: = \: 2 \: \\ \\ \sf \implies \: \: log(x ) \: - \: log(y) \: = 2 .......equation2$

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On adding equation 1 & 2 , we get;

➝ { log(x) + log(y) } + { log(x) - log(y) } = 4 + 2

➝ log(x) + log(x) + log(y) - log(y) = 6

➝ 2 log(x) = 6

➝ log(x) = 6/2

➝ log(x) = 3

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On putting value of " log(x) " in equation 2 we get;

➝ 3 - log(y) = 2

➝ log(y) = 3 - 2

➝ log(y) = 1

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• log(x) = 3
• log(y) = 1

HENCE PROVED