3 {}^(x) * 3 {}^(y ) = 1 \\ 2 {}^(2x - y) = 64 equation 1 and equation 2 find x and y​

Question

`3 {}^(x) * 3 {}^(y ) = 1 \\ 2 {}^(2x - y) = 64`

equation 1 and equation 2
find x and y​

Answer 1

Step-by-step explanation:

since x^0 =1

hi if u can't get it ask me again

Answer 2

`[2, -2]`

Explanation:

{2²ˣ ⁻ ʸ = 64

{3ˣ × 3ʸ = 1

For the second equation, it is set equal to 1, therefore you need to set it as an exponent with a zero in the exponent spot because anything raised to the zero power is 1, and according to the Product-to-Power Exponential Rule, whenever you multiply similar bases, you keep the base and add the exponents. Therefore, y and x have to be additive inverses in order for them to result in zero. We also need to make sure that for the first equation, the exponent of 2x - y is set to equal six because 2⁶ equals 64. So, those numbers would be -2 and 2, but need to be placed under the correct variables:

`{2}^(2(2) - -2) = 2^(4 + 2) = {2}^(6) = 64 \\ \\ {3}^(2) * {3}^(-2) = 9 * (1)/(9) = 1`

** They work for both when y is -2 and x is 2.

I am joyous to assist you anytime.

*** You placed this under the incorrect subject though. It should be Mathematics, NOT History.