35=5/7^u

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Solution for 35=5/7^u equation: 



35=5/7^u

We move all terms to the left:

35-(5/7^u)=0



Domain of the equation: 7^u)!=0

u!=0/1

u!=0

u∈R



We get rid of parentheses

-5/7^u+35=0

We multiply all the terms by the denominator


35*7^u-5=0

Wy multiply elements

245u-5=0

We move all terms containing u to the left, all other terms to the right

245u=5

u=5/245

u=1/49

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