Solve the log problem.

log[(a+b)/3] = (log a + log b)/3, then a/b + b/a =

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

How do we get the integer solutions ?????

Abhishek Mohapatra - 8 years, 1 month ago

it appears to be a very easy question......:) here is the probable solution....: log[(a+b)/3]=(log a+log b)/3
=>log[(a+b)/3]=log(ab)/3
=>3log[(a+b)/3]=log(ab)
=>log[{(a+b)/3}^3]=log(ab)...........taking antilog,....... =>[(a+b)^3]/[27]=ab
=>a^3+b^3+3ab(a+b)=27ab
=>so the integer solutions are [a=2 & b=4] ; [a=4 &b=-16] ; [a=4 & b=2].....BUT YOU SEE "b" CAN NOT TAKE NEGATIVE VALUE INSIDE LOGARITHM.....SO THE SOLUTION SHOULD BE.....[a=2 & b=4] AND [a=4 & b=2].......................... HOPE THIS HELPS.........:))

Sayan Chaudhuri - 8 years, 1 month ago

how did you directly deduce the integer solns from the step:=>a^3+b^3+3ab(a+b)=27ab

Bhargav Das - 8 years, 1 month ago

ha ha....!! ...i thought if anyone asks me that step by step solution.....i would crumble....!!!.....then to find the solution of a^3+b^3+3ab(a+b)=27ab i only put my graphing knowledge.........BUT EASY TO SAY YOU MAY FIND THE SOLUTION WITHOUT APPLYING YOUR BRAIN EVEN BY CLICKING HERE SIMPLY.......i only used this website to check my answer whether it is right or wrong......:-)

Sayan Chaudhuri - 8 years, 1 month ago

How do we get the integer solutions ?????

Abhishek Mohapatra - 8 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$