These posts will just feature a set of random problems I find on the net, which I will try to solve!
Interesting one is not it?
6^(x) + 6^(y) = 42
x + y = 3
Instead of taking logarithms, we will try our best to solve this the lazy way, if there is.
Alright, so let us divide 42 in such a way, we put it in exponents of 6.
We can crazily figure out:
42 is just 6(2) = 36, (+), 6(1) = 6
36 + 6 = 42.
The problem is solved:
Better representation:
6^(x) + 6^(y) = 6^(2) + 6^(1)
x = 2, y=1
2 + 1 = 3.
Thus if x = 2, y = 1
or x = 1, y = 2.
Problem 2)
Try solving the problem in a similar way I did before.
4^(x-1) + 4(3-x) = 257
Just by looking at 257 = (256 + 1)
At 256 looks like a multiple which can be expressed in terms of an even power.
4^(4) = 4 x 4 x 4 x 4 = 256
The better way to write the equation is:
4^(x-1) + 4^(3-x) = 4^(4) + 4^(0) [Because anything to power 0 is simply 1]
Note that RHS can also be inverted so two cases of x exist.
Check for which case of x the statement equals 257.
Accordingly x-1 = 4 and 3-x = 0
x = 5 , x = 3 ( x cannot be two values simultaneously)
So x - 1 = 0 and 3 - x = 4
x = 1 and x = -1.
Thus x = -1 or x = 3!
This is the perfect case!
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