(1/7)^m+2=49^m+5

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Solution for (1/7)^m+2=49^m+5 equation: 



(1/7)^m+2=49^m+5

We move all terms to the left:

(1/7)^m+2-(49^m+5)=0



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

m!=0/1

m!=0

m∈R



We add all the numbers together, and all the variables

(+1/7)^m-(49^m+5)+2=0

We get rid of parentheses

(+1/7)^m-49^m-5+2=0

We multiply all the terms by the denominator


(+1-49^m*7)^m-5*7)^m+2*7)^m=0

Wy multiply elements

-245m^2+(+1-49^m*7)^m=0

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