Logarithm proof - help: a² + b² = 6ab iff log(a + b) - log(a - b) = ½log2

[dr.trovacek]

Logarithm proof - help: a² + b² = 6ab iff log(a + b) - log(a - b) = ½log2


Prove that for positive real numbers a and b (a > b) is true that a² + b² = 6ab if and only if:
log(a + b) - log(a - b) = ½log2.

Any help solving this would be greatly appreciated. :smile:

 

[Deleted member 4993]

Prove that for positive real numbers a and b (a > b) is true that a² + b² = 6ab if and only if:
log(a + b) - log(a - b) = ½log2.

Any help solving this would be greatly appreciated. :smile:

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[stapel]

Prove that for positive real numbers a and b (a > b) is true that a² + b² = 6ab if and only if:
log(a + b) - log(a - b) = ½log2.

To prove in the one direction, you started with the log equation, applied log rules to convert the equation to "log(something) equals log (something else)", solved, and... then what?

Please be complete. Thank you!