Symbols in place of an action: If 6^x&36^2y=1 and 5^5x&25^y=1/25, then x+y=

[TreasureDragon]

I completely forgot what these types of problems are referred to as but the question is this: (I'll use & as the symbol instead of a box)

If 6^x&36^2y=1 and 5^5x&25^y=1/25, then x+y=

A -1/6
B 4/9
C -6/25
D -2/3
E -1/3

I wish I can have a starting point but I am completely lost.

 

[Deleted member 4993]

I completely forgot what these types of problems are referred to as but the question is this: (I'll use & as the symbol instead of a box)

If 6^x&36^2y=1 and 5^5x&25^y=1/25, then x+y=

A -1/6
B 4/9
C -6/25
D -2/3
E -1/3

I wish I can have a starting point but I am completely lost.

What are those?!!

Please review your post and edit.

A posted - does not make sense!!

 

[pka]

I completely forgot what these types of problems are referred to as but the question is this: (I'll use & as the symbol instead of a box)
If \(\displaystyle 6^x\cdot36^{2y}=1~\&~ 5^{5x}\cdot 25^y=\dfrac{1}{25}\text{ then } x+y=~?\)

A -1/6
B 4/9
C -6/25
D -2/3
E -1/3

I wish I can have a starting point but I am completely lost.

HINTS:
\(\displaystyle 6^x\cdot36^{2y}=6^x\cdot6^{4y}=6^{x+4y}\)

\(\displaystyle 5^{5x}\cdot 25^y=5^{5x}\cdot 5^{2y}=5^{5x+2y}\)

\(\displaystyle \dfrac{1}{25}=5^{-2}\)

 

[TreasureDragon]

HINTS:
\(\displaystyle 6^x\cdot36^{2y}=6^x\cdot6^{4y}=6^{x+4y}\)

\(\displaystyle 5^{5x}\cdot 25^y=5^{5x}\cdot 5^{2y}=5^{5x+2y}\)

\(\displaystyle \dfrac{1}{25}=5^{-2}\)

So the symbols represent multiplication?

Assuming this I have concluded with:

x+4y=0 and 5x+2y=-2 and got x= -4/9 and y= 1/9, getting the answer of E. Is this correct?

 

[Deleted member 4993]

So the symbols represent multiplication?

Assuming this I have concluded with:

x+4y=0 and 5x+2y=-2 and got x= -4/9 and y= 1/9, getting the answer of E. Is this correct?

What do you think?

Put those values in your original equations and check!