Question

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its true or false pls give the ans and the WORKING ​

Answer 1

Answer:

If ( x , y ) is the answer of the given equation ( 2 x ) log 2 = ( 3 y ) log 3 and 3 log x = 2 log y then x is equal to?

Explanation:

Taking the logarithm on both sides (the first equation) we get

log ( 2 ) ( log ( 2 ) + log ( x ) ) = log ( 3 ) ( log ( 3 ) + log ( y ) )

Doing the same with the second equation:

log ( y ) = log ( x ) ⋅ log ( 3 )/ log ( 2 )

Substituting

a = log ( x )

we get

log ( x ) log 2 ( 2 ) − log 2 ( 3 )/ log ( 2 ) = − ( log 2 ( 2 ) − log 2 ( 3 ) )

so

a = − log ( 2 )

log ( x ) = log ( 2^{-1}2

−1

)

x = 1 /2

In the first equation we get

1 = 3 y

y = 1 /3

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Hope this helps.

Answer 2

From 1st equation:-

(2x)^(log2) = (3y)^(log3)

Taking log on both sides

→ (log2) . (log2x) = (log3) . (log3y)

→ (log2) . (log 2 + logx) = (log3) . (log3 + logy) ----- (i)

From 2nd equation:-

3^(logx) = 2^(logy)

Taking log on both sides

→ (logx) . (log3) = (logy) . (log2) ------- (ii)

Let log2 be a, log3 be b, logx be m, and logy be n

So, we can write (i) as :-

a(a + m) = b(b + n)

And, (ii) can be written as :-

bm = an

→ n = (bm)/a ----- (iii)

Putting (iii) in (i) :-

a(a + m) = b(b + n)

→ a² + am = b² + bn

→ a² + am = b² + b{(bm)/a}

→ a² + am = b² + (b²m)/a

→ a² - b² = (b²m)/a - am

→ a² - b² = (b²m - a²m)/a

→ a(a² - b²) = m(b² - a²)

→ a(a + b)(a - b) = m(b - a)(b + a)

→ a(a - b) = m(b - a)

→ a(a - b) = -m(a - b)

→ a = -m

So,

log 2 = -1 * log x

→ log 2 = log (x)^(-1)

→ 2 = (x)^(-1)

→ x = 1/2

From (iii) :-

n = (bm)/a

→ log y = [ {log (1/2)} . {log (3)} ] / [log 2]

→ log y = [ {log (2)^(-1)} . {log (3)} ] / [log 2]

→ log y = [ -1 . log 2 . log 3] / [log 2]

→ log y = -1. log 3

→ log y = log (3)^(-1)

→ y = (3)^(-1)

→ y = 1/3

So,

[ (x)^(-1) + (y)^(-1) ]

→ [ (1/2)^(-1) + (1/3)^(-1) ]

→ [ 2 + 3 ]

→ 5

But is is given that [ (x)^(-1) + (y)^(-1) ] = 4. So Ans is

False.

Ans. False