How do you solve #log_2(2x)=-0.65#?

Question

3037 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-06T22:10:04

#x=0.319color(white)(888)#3 .d.p.

By the law of logarithms:

If:

#y=log_ba#

Then:

#b^y=a#

Since: #y=log_ba#

#b^(log_ba)=a#

Using this idea:

#log_2(2x)=-0.65#

#2^(log_2(2x))=2^(-0.65)#

#2x=2^(-0.65)#

Dividing by #2#:

#x=2^(-0.65)/2=0.319color(white)(888)#3 .d.p.