Question

if log (a^3-b^3)-log 3=3/2 (log a+lob b) then find the value of (a/b)^3 +(b/a)^3

Answer 1

$log(a^3-b^3)-log3=\frac{3}{2}(loga+logb)$

$2[log(a^3-b^3)-log3]=3(loga+logb)$

$2log(a^3-b^3)-2log3=3(loga+logb)$

we know,
$logM+logN=logMN\\\\AlogM=logM^A$

$log(a^3-b^3)^2-log3^2=3logab$

$log(a^3-b^3)^2-log9=log(ab)^3$

$log(a^3-b^3)^2-log(ab)^3=log9$

we also know, $logA-logB=log\frac{A}{B}$

$log\frac{(a^3-b^3)^2}{(ab)^3}=log9$

removing log from both sides,

$\frac{(a^3-b^3)^2}{(ab)^3}=9$

$\frac{a^6+b^6-2a^3b^3}{a^3b^3}=9$


$\frac{a^6}{a^3b^3}+\frac{b^6}{a^3b^3}-\frac{2a^3b^3}{a^3b^3}=9$

$\frac{a^3}{b^3}+\frac{b^3}{a^3}-2=9$

$(a/b)^3+(b/a)^3=9+2=11$

Answer 2

Hi ,

It is given that ,

log(a³-b³)-log3 = 3/2[logs+logb]

=> log[(a³-b³)/3] = 3/2log(ab)

***********************************

By Logarithmic rules :

1 ) log m - log n = log m/n

2 ) log m + log n = log ( mn )

3 ) n log m = log mⁿ

****************************************

=> 2log[ (a³-b³)/3] = 3log(ab)

=> log[(a³-b³)/3]² = log(ab)³

Remove log both sides of the

equation , we get

(a³- b³)²/3² = (ab)³

=> (a³-b³)² = 9a³b³

=> (a³)² + (b³)² - 2a³b³ = 9a³b³

=> (a³)² + ( b³ )² = 11a³b³

=> Divide each term with a³b³,

both sides of the equation

[(a³)²/(a³b³)+ [(b³)²/(a³b³)] = 11a³b³/a³b³

=> (a/b)³ + ( b/a)³ = 11

I hope this helps you.

: )