Solving Exponential Equation @KasyannoEZMath

This tutorial video explains how to solve the given exponential equation.

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[peterbyrne]

Hi, I like your way of solving such kind of exponential equations Thank you..

[lightyagami]

Can be done mostly mentally (I did).

"Cast out" fours on the RHS by division. You can cast out three 4s. Because 1088 = 17 . 4^3

Divide both sides by this power of 4.

So you get 4^(x-3) + 4^(y-3) = 17

Note that the RHS is odd, and therefore the LHS must also be odd. For any positive exponent, each term is even, so you need a zero exponent for one term to be odd. The solutions will therefore have either y - 3 = 0 or x - 3 = 0.

In the first case, y = 3 so 4^(x-3) + 4^0 = 17, implying that 4^(x-3) = 16 = 4^2 so x - 3 = 2 so x = 5.

The cases are symmetrical, so in the second case, x = 3, y = 5.

The solutions are therefore (x, y) = (3, 5) or (5, 3).

[thorinpalladino]

This method only works for rational numbers, so if you are going to limit your solutions to that set you should state it in the intro.
if 4^x =1087 and 4^y =1 you still have 4^x + 4^y =1088. x= log (base 4) 1087, y=0. This is just one of an infinite number of possible irrational solutions.

[jjg]

2^10 =1024, 2^6=64;
1024+64 = 1088;
2^2x = 1024 or 2^10;
2x = 10
x = 5

2^2y = 64 or 2^6;
2y = 6
y = 3
thus x = 5 or 3
y =3 or 5

[nikolayguzman]

There are another way: 4^x=2^2x: 4^y=2^2y; 2^10=1024; 2^6=64; 1024+64=1088; 2^2x=2^10 2x=10; x=5; 2y=6 y=3; This way is shorter.

[SL-uppl]

I don’t know maths well but may I do this way: 4^x + 4^y = 4^5 +4^3, then x + y = 5 + 3 so x = 5 or 3; y = 3 or 5. How did I find out 1088 = 4^5 + 4^3? I just use the calculator 4 x 4… until the 5th time, the answer is 1024, then 1088 - 1024 = 64, it is easy to tell 64 = 4^3 or 4 x 4… again if you want.

[williamloud]

How does anybody come up with these tricks? What were the logical steps that led to add (y-y) to an exponent?

[kfjfkeofitorhf]

4^5+4^3=1088
(2^2)^5+(2^2)^3=
2^10+2^6=1088
1024+64=1088