How do you solve #log_2(2x)=log_2 100#?

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Answer 1 · 2018-07-12T15:28:35

#color(blue)(x=50)#

If:

#log_a(b)=log_a(c)=>b=c#

Hence:

#log_2(2x)=log_2(100)#

#2x=100=>x=100/2=50#

Answer 2 · 2018-07-12T22:07:27

#x=50#

Since our bases are the same, we can essentially cancel them out and be left with

#2x=100#

By dividing both sides by #2#, this easily simplifies to

#x=50#

Hope this helps!