Sure, Two Times Three Is Six

$2^x = 6 \Longrightarrow 2 = 6^\frac{1}{x} ... (1)$

$3^y = 6 \Longrightarrow 3 = 6^\frac{1}{y} ... (2)$

Multiplying the RHS's of eqs (1) and (2):

$6^1 = 6^{\frac{1}{x} + \frac{1}{y} }$

So, $1 = \frac{1}{x} + \frac{1}{y} \Longrightarrow \boxed{x+y=xy}$

$2^x = 3^y = 6\quad \Rightarrow x\log{2} = y\log{3} \quad \Rightarrow \dfrac {x}{y} = \dfrac {\log{3}}{\log{2}}$

$x\log{2} = \log{6} = \log{2}+\log{3}\quad \Rightarrow x = 1 + \dfrac {\log{3}}{\log{2}}$

$\Rightarrow x = 1 + \dfrac {x}{y}\quad \Rightarrow xy = y+x \quad \Rightarrow \boxed {x+y=xy}$

Firstly, multiply each of the exponents by ${x}$ such that $6^{x}$ = $2^{x}$ $\times$ $3^{x}$ = 6 $\times$ $3^{x}$ = $3^{xy}$ ( 1). Then create another equation by multiplying the exponents of the original equation by ${y}$ such that: $6^{y}$ = $2^{y}$ $\times$ $3^{y}$ = 6 $\times$ $2^{y}$ = $2^{xy}$ (2 ). Multiplying equations ( 1) and (2 ) together gives: $6^{x+y}$ = $6^{xy}$ $\therefore$ ${x+y}$ = ${xy}$

I did it in my head. The only way the fourth option works is if x any y are both 1 or -1. Not possible because 2^x =6

y-x = xy also eliminated as x must be larger than y and this would cause xy to be a negative number. Both exponents must be positive to satisfy the equation.

x-y=xy also no good. Both exponents must be greater than 1 and both positive to satisfy the equation. Since x>y >1 xy=x-y has no solution.

Since this is a legit problem not a trick question the only remaining solution x+y=xy is therefore the correct one.

Not a rigorous or elegant solution I admit but sound reasoning producing the correct answer. Took a lot longer to type than to work out.

This approach was workable only because it is a multiple choice question. If the question had been asked give the relationship between x and y I would have had to find a piece of paper.

$2^{ x }=6=>x=\log _{ 2 } 6=\log _{ 2 } (2*3)=\log _{ 2 } 2+\log _{ 2 } 3=1+\log _{ 2 } 3.\quad \\ 3^{ y }=6=>y=\log _{ 3 } 6=\log _{ 3 } (3*2)=\log _{ 3 } 3+\log _{ 3 } 2=1+\log _{ 3 } 2.\quad \\ xy=(1+\log _{ 2 } 3)(1+\log _{ 3 } 2)=1+\log _{ 2 } 3+\log _{ 3 } 2+\log _{ 2 } 3*\log _{ 3 } 2=2+\log _{ 2 } 3+\log _{ 3 } 2.\quad \\ I\quad used\quad the\quad fact\quad that\quad \log _{ a } b*\log _{ b } a=1.\quad \\ x+y=1+\log _{ 2 } 3+1+\log _{ 3 } 2=2+\log _{ 2 } 3+\log _{ 3 } 2.\\ The\quad conclusion\quad is:\quad x+y=xy$

• $2^{x-1}=3$ since $3^{y}=6$ $(2^{x-1})^{y}= 6$ $2^{xy-y}=2^{x}=6$ Equate the powers xy-y=x x+y=xy

If one is trying to find the correct one out of the alternatives given then x+y=xy is correct. However, if one substitutes these values in the given equation, it will not hold good as 2^2 is 4 and 3^2 is 9 both are quite different and also not equal to 6. Hence there appears some error in the given equation itself.

$(2^{x})^{y}=6^y$ and $(3^{y})^{x}=6^x$ . So $2^{xy} 3^{xy}=6^{x} 6^{y}$ , that is, $6^{xy}=6^{x+y}=$ . Equate exponents.

Let us assume that $2^{x} = 3^{y} = 6^{1} = k$

So

$2^{1} = k^{1/x} and 3^{1} = k^{1/y}$

Now,

$6 = 2^{1} \times 3^{1}$

So

$\frac{1}{x}$ + $\frac{1}{y}$ = 1

So

$x + y = x \times y$

The other authors have good explanations of the formal mathematical solution. This solution discusses answering the multiple choice question quickly, such as on a timed, standardized exam such as an SAT or GRE.

First, by looking at the equation, we know that x and y must be positive real numbers, both greater than 1, with x > y. Therefore the statements y - x = xy; xy = x / y; and x - y = xy must all be false, leaving x + y = xy the best choice.

y - x will be negative, xy must be positive.

For two numbers greater than 1, x / y must be closer to 1 than xy (this logic holds true for y / x = xy being false as well - it does not matter whether "closer to 1" is greater or less than 1, it is still less than xy).

Likewise, x - y must be closer to 0 than xy (again, same logic for y - x = xy being false).

So, on an exam, select x + y = xy, possibly marking it for a more rigorous check at the end if time permits.

So I used process of elimination. X and y must both be greater than 1 since 2^x and 3^y both equal six. Therefore, xy is greater than either x or y so subtracting these two variables in any order does not work since that will give a number less than either of them. Also, xy=x/y does not work since y would have to be one. That leaves only one answer.

I'm not sure with my solution but I got it right by doing this : 2^x = 6 -> log6 base 2 = x (1) 3^y = 6 -> log6 base 3 = y (2)

Then, we need to change the bases of eq 1 and 2 so they have same base, which is 10 [that's what I want XD] so [using a theorem wherein I forgot what it is called XD ]

x = log6/log2 and y = log6/log3

try adding them and you'll see that it is true that

log6/log2 + log6/log3 = log6(log3+log2)/log2*log3

note that log3+log2 = log6 [using a rule in addition of logs]

so log6(log3+log2) = log6*log6

try muliplying the two of them and you'll see that

log6/log2 * log6/log3 = log6(log6)/log2*log3

which is the same as the first one :-D

Hope it helped somebody to understand the problem :-D

I m really Brilliant

2^x=6 and 3^y=6, therefor x>1 and y>1; xy>x/y (xy=x/y is not correct); x-y<xy (x-y=xy is not correct); 2<3, therefor x>y; y-x<0 and xy>0 (y-x=xy is not correct); x+y>x and x+y>y and xy>x and xy>y, so x+y=xy has to be correct answer (assuming that one answer is correct)

without solving, option x+y=xy is the only simetric equation and the original equations cannot bring a-simetric ones

2^x=6 => 2^(x-1)=3 and
3^x=6 it result that
2^((x-1) y)=6 it result that
2^((x-1)
y)=2^x therefore (x-1) y=x and x y=x+y