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

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 get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

-343m^2-35m+14m+1=0

We add all the numbers together, and all the variables

-343m^2-21m+1=0

a = -343; b = -21; c = +1;
Δ = b2-4ac
Δ = -212-4·(-343)·1
Δ = 1813
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1813}=\sqrt{49*37}=\sqrt{49}*\sqrt{37}=7\sqrt{37}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-7\sqrt{37}}{2*-343}=\frac{21-7\sqrt{37}}{-686}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+7\sqrt{37}}{2*-343}=\frac{21+7\sqrt{37}}{-686}
$