How do you solve for x in #log_2x-log_2 3=5#?

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Answer 1 · 2017-02-01T09:09:04

#x=96#

Let's first combine the left side, using the rule that #log_x(a)-log_x(b)=log_x(a/b)#:

#log_2(x)-log_2(3)=5#

#log_2(x/3)=5#

We can now set these as exponents to a base of 2 (to eliminate the log):

#2^(log_2(x/3))=2^5#

On the left side, this has the effect of taking the inverse function of the log and so cancels out, leaving:

#x/3=32#

#x=96#

And now let's check it:

#log_2(96)-log_2(3)=5=>6.585-1.585=5#

(the logs are rounded to 3 decimal places)