Manipulating Exponent

$3^x = 2^{2015} => 3^{xy} = 2^{2015y}$

But, $2^{2015y} = 81$

$=> 3^{xy} = 81 => 3^{xy} = 3^4$ $=> xy = 4$

3^^x=2^^2015 implies x=2015log(2)/Log(3) 2^^2015y=81 implies y=4 log(3)/2015 log(2) so xy=2015 log(2) * 4log(3)/log(3)*2015 log(2) after cancelling out terms XY becomes 4

3= 2^2015 /3^x-1, 3= 2^2015y/3^3, Thus by Cross Multiplication 3^3 * 2^2015= 3^(x-1) * 2^2015y, Thus If x-1=3, x=4 and y=1, Thus xy=4

3^x=2^2015 and 2^2015y=81.So 3^xy=81,3^xy=3^4 so xy=4

We have to take natural log in both these equations 3^x = 2^2015 & 2^2015y = 81

ln3^x = ln 2^2015

xln3 = 2015 ln2

x = 2015 ln2/ln3

ln 2^2015y = ln 81

2015yln2 = ln 81

y = ln 81/2015 ln 2

Now xy = [2015ln2/ln3 ] * [ln81/2015ln2 ]

Now the 2015 ln 2 gets cancelled out so we are left with

xy= ln81/ln3 = 4

3^x = 2^2015 2^(2015y) = 81 3^(xy) =81 xy=4

Because my lucky number is four

2^2015y = 3^xy 81 = 3⁴ Hence xy = 4

x=4

y=1 xy=4

Taking log both sides of both equation and then simply simplifying... Xlog3 = 2015log2 , 2015ylog2 = 4log3. , we get xy = 4

x=2015(log2)base3. y=4/2015(log3)base2. multiplying. (log2)base3 (log3)base2=1.(log property); so x y=4...

3 POWER 4 (3^4)= 2 POWER 2015 Y, 3 POWER x (3^x)=2 POWER 2015 , SO X=4,Y=1 X Y=4 1=4

3^4= 2^(2015y), 3^x=2^(2015); 3=2^(2015y/4), 3=2^(2015/x); 2015y/4=2015/x; y/4=1/x xy=4

3^x = 2^2015

(3^x)^y = (2^2015)^y

3^xy = 2^2015y = 81 = 3^4

xy = 4

its a matter of understanding properties of exponents. Substitute 3^X to 2^2015 in in 2nd equation. making it 3^xy = 81. then 3^xy = 3^4....therefore xy = 4......

3^x = 2^2015 2^2015y = 81 (y != 0, because 2^2015y = 81 => 1 = 81) (y != 1, because 2^2015 = 81 => 2^7 = 128, 128 > 81 => 2^2015 > 81 (2^x < 2^(x+1)) (2^2015)^y = 3^4 2^2015 = 3^(4/y) 3^x = 3^(4/y) x = 4/y xy = 4

Logarithm can be used too, but could be tedious