How do you solve #log_2(x-6)=5#?

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Answer 1 · 2016-04-11T18:37:19

#38#

Raise both sides to #2#,

#2^(log_2(x-6)) = 2^5#
#x - 6 = 32#

Add #6# to both sides to cancel out the subtraction,

#x = 38#

Answer 2 · 2016-04-11T18:43:42

The same thing but with a slightly different approach.

#x=38#

For this example I am using #log_10# so that you can check it on a calculator.

Consider #log_10(x)=3#

This is another way of writing #10^3=x#

So #x=1000#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using the above method

#log_2(x-6)=5 " "-> " " 2^5=x-6#

#=>x=2^5+6#

#=> x= 32+6=38#