'Loggic' equation

Algebra Level pending

log ⁡ a + log ⁡ b = log ⁡ ( a + b ) \log a+\log b=\log(a+b)

Which option does not satisfy the above equation?

b = a a − 1 b=\dfrac{a}{a-1} a = b b − 1 a=\dfrac{b}{b-1} a = b = 2 a=b=2 a = b = 1 a=b=1

Dmitri Ivanovich by Dmitri Ivanovich , 19, Japan

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1 solution

We want a b ≠ a + b ab \neq a+b . So we eliminate the options which satisfy a b = a + b ab=a+b .

If a = b a=b , solving we get a = b = 2 a=b=2 .

Simplifying the options a = b b − 1 a=\dfrac b {b-1} and b = a a − 1 b=\dfrac a {a-1} we get a b = a + b ab=a+b .

The only option that does not satisfy the equation is a = b = 1 a=b=1 .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

@Dmitri Ivanovich and @A Former Brilliant Member , There is not only a = b = 1 does not satisfy the logarithmic equation. a = b = 0 also does not satisfy the equation above. \text { There is not only } a = b = 1 \text { does not satisfy the logarithmic equation. } a = b = 0 \text { also does not satisfy the equation above. }

. . - 3 months, 2 weeks ago

Hmm, nice solutioj, ypu have jussed missed to prove that a = b = 1 a = b = 1 is wrong which can be done by mere substitution. Plz update your solution accordingly and receive and upvote! Man, I was pedantic there XD. (≧▽≦)

Ashish Menon - 5 years ago

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