How do you simplify #sqrt2^109 * sqrtx^306 * sqrtx^11#?

Question

1547 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-25T11:08:20

#2^54x^158sqrt(2x)#

We have #(sqrt2)^109 xx (sqrtx)^306 xx (sqrtx)^11#

We can write this as #2^(109/2) xx x^(306/2) xx x^(11/2)#

We can use the rule #x^a xx a^b = x^(a+b)# to combine the #x# terms:

#2^(109/2) xx x^(306/2+11/2)#

#2^(109/2) xx x^(317/2)#

#2^54 xx 2^(1/2) xx x^158 xx x^(1/2)#

#2^54x^158sqrt(2x)#

If there is a desire to evaluate #2^54#, it's roughly #1.8 xx 10^16#