+1926 This problem looks really complicated and a nightmare to solve, but it actually is quite simple!
First, we have to factor in the 4m-2 into the parenthesis, getting us \(\frac{{1}^{4m-2}}{{49}^{4m-2}} = {7}^{-10m-2}\).
This still looks far off from our goal of finding m, but notice that 49 is just 7^2 and the numerator is just 1. This means that we just have \(\frac{1}{{7}^{2(4m-2)}}\).
Alright, now we're closer to comparable levels. However, we have once term in the denominator and one in the numerator. What should we do about that?
Well, the answer lies in the fact that \({a}^{-1} = \frac{1}{a}\). If we factor out -1 from the power in the right side of the equation, we have \({7}^{-(10m+2)}\), meaning that we technically just have the final equation \(\frac{1}{{7}^{8m-4}} = \frac{1}{{7}^{10m+2}}\). In order for the equation to be satisfied, we must have \({7}^{8m-4} = {7}^{10m+2}\), meaning that \(8m - 4 =10m + 2\).
Combining like terms gets us \(-2m = 6\) so \(m = -3\)!
The key to this problem is to realize that we have to kind of force our way to a similar expression on both sides so that we could write an equation. I hope I answered your question correctly and clearly!
Thanks!
