Question

if log (a-b÷5)=1÷2(log a+log b) then find a÷b+b÷a​

Answer 1

Answer:-

This is the required solution.

Given,

$\sf log( \frac{a - b}{5} ) = \frac{1}{2} ( log(a) + log(b) )$

$\sf \implies log( \frac{a - b}{5} ) = \frac{1}{2} log(ab)$

$\sf \implies log( \frac{a - b}{5} ) = log( \sqrt{ ab})$

Removing log from both side, we get,

$\sf \implies \frac{a - b}{5} = \sqrt{ ab}$

$\sf \implies a - b = 5\sqrt{ ab}$

Now, squaring both side, we get,

$\sf \implies (a - b)^{2} = (5\sqrt{ ab})^{2}$

$\sf \implies {a}^{2} + {b}^{2} - 2ab = 25 ab$

$\sf \implies {a}^{2} + {b}^{2} = 27 ab \: ...(i)$

Now,

$\sf \frac{a}{b} + \frac{b}{a}$

$\sf = \frac{ {a}^{2} + {b}^{2} }{ab}$

From (i), we can say that a²+b²=27 ab

So,

$\sf \frac{ {a}^{2} + {b}^{2} }{ab}$

$\sf = \frac{27 \cancel{ab}}{ \cancel{ab}}$

$\sf = 27$

Hence,

$\boxed{ \sf \frac{a}{b} + \frac{b}{a} = 27 }$

Which is the required answer.

Answer 2

Answer:

This is the required solution.

Given,

\sf log( \frac{a - b}{5} ) = \frac{1}{2} ( log(a) + log(b) )log(

5

a−b

)=

2

1

(log(a)+log(b))

\sf \implies log( \frac{a - b}{5} ) = \frac{1}{2} log(ab)⟹log(

5

a−b

)=

2

1

log(ab)

\sf \implies log( \frac{a - b}{5} ) = log( \sqrt{ ab})⟹log(

5

a−b

)=log(

ab

)

Removing log from both side, we get,

\sf \implies \frac{a - b}{5} = \sqrt{ ab}⟹

5

a−b

=

ab

\sf \implies a - b = 5\sqrt{ ab}⟹a−b=5

ab

Now, squaring both side, we get,

\sf \implies (a - b)^{2} = (5\sqrt{ ab})^{2}⟹(a−b)

2

=(5

ab

)

2

\sf \implies {a}^{2} + {b}^{2} - 2ab = 25 ab⟹a

2

+b

2

−2ab=25ab

\sf \implies {a}^{2} + {b}^{2} = 27 ab \: ...(i)⟹a

2

+b

2

=27ab...(i)

Now,

\sf \frac{a}{b} + \frac{b}{a}

b

a

+

a

b

\sf = \frac{ {a}^{2} + {b}^{2} }{ab}=

ab

a

2

+b

2

From (i), we can say that a²+b²=27 ab

So,

\sf \frac{ {a}^{2} + {b}^{2} }{ab}

ab

a

2

+b

2

\sf = \frac{27 \cancel{ab}}{ \cancel{ab}}=

ab

27

ab

\sf = 27=27

Hence,

\boxed{ \sf \frac{a}{b} + \frac{b}{a} = 27 }

b

a

+

a

b

=27

Which is the required answer.