How do you solve #Log_2 32 = x#?

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Answer 1 · 2016-03-09T21:16:37

#x=5#

The logarithmic expression can be exponentiated with a base of #2#:

#2^(log_2 32)=2^x#

The #2^x# and #log_2 x# functions are inverses, which means that they undo one another, so we obtain the equation:

#32=2^x#

We can write #32# as a power of #2#:

#2^5=2^x#

Since the bases are equal, we know their powers are also equal, giving:

#x=5#

Answer 2 · 2016-03-09T21:18:12

#x=5#

Since #32=2^5#, we see that

#log_2 2^5=x#

Using the logarithm rule:

#log(a^b)=b*log(a)#

Giving the equation:

#5*log_2 2=x#

Since #log_a a=1#,

#x=5#