Question

if log (a- b ) /5 = 1/2 (log a + log b ) show that A^2 + b^2 =27 ab
​

Answer 1

Question :

if log (a- b ) /5 = 1/2 (log a + log b ) show that a² + b² =27 ab

Given :

log (a - b)/5 = 1/2 (loga + logb)

To show :

a² + b² = 27ab

Solution :

Let's start with what is given :

⇒  log(a - b)/5 = 1/2 (log a + log b)

⇒  2 log (a - b)/5 = log a + log b

⇒  log { (a - b)/5 }² = log(ab)    [∵ n logₐ b = b logₐⁿ]

⇒  {(a - b)/5}² = ab   [Taking away log from both sides]

⇒  (a² - 2ab + b²)/25 = ab

⇒  a² - 2ab + b² = 25ab

⇒  a² + b² = 25ab + 2ab

⇒  a² + b² = 27ab    [Showed]

Therefore,

If log (a - b )/5 = 1/2 (log a + log b ), then a²+ b² = 27 ab   [showed]

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

• If $\sf{log\left(\dfrac{a-b}{5}\right)}=\sf{\dfrac{1}{2}\:\left(log\:a\:+\:log\:b\right)}$  show that a² + b² =27ab

★═════════════════★  

♣ ɢɪᴠᴇɴ :

• $\sf{log\left(\dfrac{a-b}{5}\right)}=\sf{\dfrac{1}{2}\:\left(log\:a\:+\:log\:b\right)}$

★═════════════════★  

♣ ᴛᴏ ꜱʜᴏᴡ :

• a² + b² =27ab

★═════════════════★  

♣ ᴀɴꜱᴡᴇʀ :

$\sf{log\left(\dfrac{a-b}{5}\right)}=\sf{\dfrac{1}{2}\:\left(log\:a\:+\:log\:b\right)}$

Multiplying both sides with 2 :

$\sf{2\times log\left(\dfrac{a-b}{5}\right)}=\sf{2\left[\dfrac{1}{2}\left(log\:a+log\:b\right)\right]}$

$\sf{2\: log\left(\dfrac{a-b}{5}\right)}=\sf{\left(log\:a\:+\:log\:b\right)}$

Using logarithm property :  $\sf{n\log_a\: b = b\:\log_a \:^n}$

$\sf{\log \left(\dfrac{a-b}{5}\right)^{2}=\log (a b)}$

Taking log on both sides :

$\sf{\left(\dfrac{a-b}{5}\right)^{2}=a b}$

$\sf{\dfrac{(a-b)^{2}}{25}=a b}$

Cross multiply :

$\sf{(a-b)^{2}=25 a b}$

We know (a - b)² is an algebraic identity and (a - b)² = a² + b² - 2ab

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

Adding 2ab on both sides :

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

$\large\boxed{\sf{a^2+b^2=27ab}}$