solve it as it is hard question
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2^x= k =>2= k^1/x
3^y=k=>3=k^1/y
6^-z=k=>6=k^-1/z
k^1/x × k^1/y= 2×3
k^1/x+1/y= 6
k^1/x+1/y= k^-1/z
1/x +1/y = -1/z
1/x +1/y+ 1/z= 0
hope this helps you.
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3^y=k=>3=k^1/y
6^-z=k=>6=k^-1/z
k^1/x × k^1/y= 2×3
k^1/x+1/y= 6
k^1/x+1/y= k^-1/z
1/x +1/y = -1/z
1/x +1/y+ 1/z= 0
hope this helps you.
Mark me brainliest
sarika18082002:
I can't understand plz give pic of answer
hi
plz
give pic of answer
i have not option to edit this answer
repost your question
kk
see I have posted
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