Kann man hier den Logarithmus ausdrücken. Wenn ja zeige es auf:

Question

Bsp.)

1/4* [lga-2*lgb+3lgc-lg(a-b)]

= 1/4 [(lga-2)+lg(b)*3lgc-lg(a-b)]

=1/4*[(lga-2)+((lg(b)*lgc))^1/3]/lg(a-b)

=1/4*[(lga-2)+lg(b)^1/3*lg(c)^1/3)]/lg(a-b)

=$$\sqrt [ 4 ]{ \frac { bc³ }{ b²*(a-b)} } $$

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Answer 1

1/4·(LOG(a) - 2·LOG(b) + 3·LOG(c) - LOG(a - b))

= 1/4·(LOG(a) - LOG(b^2) + LOG(c^3) - LOG(a - b))

= 1/4·LOG(a·c^3/(b^2·(a - b)))