Question

D%5E%7B2x%20-%20y%7D%20%20%20%3D%2064" id="TexFormula1" title="3 {}^{x} \times 3 {}^{y } = 1 \\ 2 {}^{2x - y} = 64" alt="3 {}^{x} \times 3 {}^{y } = 1 \\ 2 {}^{2x - y} = 64" align="absmiddle" class="latex-formula">
equation 1 and equation 2
find x and y​

Answer 1

Explanation:

since x^0 =1

hi if u can't get it ask me again

Answer 2

Answer:

$[2, -2]$

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:

${2}^{2(2) - -2} = 2^{4 + 2} = {2}^{6} = 64 \\ \\ {3}^{2} \times {3}^{-2} = 9 \times \frac{1}{9} = 1$

** 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.