Solve 7^x + 343/7^x = 56

You just need to replace the 7^x with something like k or a. This makes urs simplification method much more easier.

So the solution is:

7^x + $\frac{343}{7^x}$ = 56

let 7^x = k

so, k + $\frac{343}{k}$ = 56
or, k² - 56k + 343
or, k² - (7 + 49)k + 343 = 0
or, k² - 7x - 49k + 343 = 0
or, k(k - 7) - 49(k - 7) = 0
or, (k - 7)(k - 49) = 0

Either, k = 7
so, 7^x = 7¹
so, x = 1

OR, k = 49
so, 7^x = 7²
so, x = 2

Hence, x = 1, 2

Video explanation: