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.
z has to satisfy \[ \large x^xy^y=z^z \] ?
yes
that is \[ \Large (2^{12}3^6)^{2^{12}3^6}(2^83^8)^{2^83^8}=z^z \] ?
yes
I wonder if logarithms would help here...
I am not sure, I haven't learn log yet
But you're comfortable with basic exponent properties, yes?
yes
Ok, start with what helder_edwin posted and simplify that using rules for exponents and try to get all your bases as prime numbers.
Then you can group the bases together.
can you show me an example?
Alright.. \[\large (2^{12}3^6)^{2^{12}3^6} (2^83^8)^{2^83^8}\] \[\large \rightarrow 2^{12\cdot2^{12}\cdot3^6}\cdot3^{6\cdot2^{12}\cdot3^6}\cdot 2^{8\cdot2^8\cdot3^8}\cdot3^{8\cdot2^{8}\cdot3^8}\]
With a power-to-a-power, you multiply exponents.
|dw:1350088446554:dw| ?
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