How do you solve #15+2log_2x=31#?

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Answer 1 · 2016-09-07T14:58:02

#x = 256#

We have: #15 + 2 log_(2)(x) = 31#

First, let's subtract #15# from both sides of the equation:

#=> 2 log_(2)(x) = 16#

Then, let's divide both sides by #2#:

#=> log_(2)(x) = 8#

Now, using the laws of logarithms:

#=> x = 2^(8)#

#=> x = 256#

Therefore, the solution to the equation is #x = 256#.

Answer 2 · 2016-09-07T15:00:04

#x=256#

#15+2log_2 x=31#

#hArr2log_2 x=31-15=16# or

#log_2=8#

Hence #x=2^8=256#