Math, asked by Dasham10, 1 year ago

If 2^a = 3^b = 6^-c, then prove that 1/a + 1/b + 1/c = 0.

Answers

Answered by SoulTushar

please see image for soln.

2nd method

2^a=3^b=6^(-c)=k

2^a=k
k^(1/a)=2

similarly
k^(1/b)=3
& k^(-1/c)=6

and since 2×3=6

[k^(1/a)]×[k^(1/b)]=k^(-1/c)
k^[(1/a)+(1/b)]=k^(-1/c)
since bases are same thats why powers will be equal
therefore, (1/a)+(1/b)= -1/c
(1/a)+(1/b)+(1/c)=0
hence proved

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Dasham10: Thanks for you answer but i havent been taught about log. I am in IX. So, there should be another way to solve this. Thanks once again!..

SoulTushar: no problem bro....I just edited my answer and given 2nd method, you can look at that

Dasham10: Thankyou so much!

SoulTushar: You're welcome

SoulTushar: Bro can you mark my answer as Brainliest answer if it served your purpose.

Dasham10: The option to mark brainliest is not coming

Dasham10: Probably because only one ans has been given

SoulTushar: okay bro no problem

Answered by Aditiiiiiiiiiii

Answer:

1/a +1/b +1/c = 0

Step-by-step explanation:

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