How do you evaluate #sqrt(log2* 20-log16) #?

Question

1240 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-09T14:32:44

The expression can be evaluated to #2.19#.

#=>sqrt(20log2 - log16)#

Use the rule #alogn = logn^a#.

#=>sqrt(log(2^20) - log16)#

#=> sqrt(log1,048,576 - log16)#

Now, use the rule #loga - logn = log(a/n)#.

#=>sqrt(log(65536))#

This can be evaluated, but it cannot be simplified to exact form.

We have

#=>2.19#

Hopefully this helps!