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Mathematics 12 Online
OpenStudy (anonymous):

If x=(2^12)(3^6) and y=(2^8)(3^8), what integer z satisfies (x^x)(y^y)=z^z? Write the answer as a product of powers of primes.

OpenStudy (helder_edwin):

z has to satisfy \[ \large x^xy^y=z^z \] ?

OpenStudy (anonymous):

yes

OpenStudy (helder_edwin):

that is \[ \Large (2^{12}3^6)^{2^{12}3^6}(2^83^8)^{2^83^8}=z^z \] ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

I wonder if logarithms would help here...

OpenStudy (anonymous):

I am not sure, I haven't learn log yet

OpenStudy (anonymous):

But you're comfortable with basic exponent properties, yes?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Ok, start with what helder_edwin posted and simplify that using rules for exponents and try to get all your bases as prime numbers.

OpenStudy (anonymous):

Then you can group the bases together.

OpenStudy (anonymous):

can you show me an example?

OpenStudy (anonymous):

Alright.. \[\large (2^{12}3^6)^{2^{12}3^6} (2^83^8)^{2^83^8}\] \[\large \rightarrow 2^{12\cdot2^{12}\cdot3^6}\cdot3^{6\cdot2^{12}\cdot3^6}\cdot 2^{8\cdot2^8\cdot3^8}\cdot3^{8\cdot2^{8}\cdot3^8}\]

OpenStudy (anonymous):

With a power-to-a-power, you multiply exponents.

OpenStudy (anonymous):

|dw:1350088446554:dw| ?

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