How do you solve for x in #-5 = log _2 x#?

1 Answer
Rachel
Mar 16, 2016

#x=1/32#

Explanation:

We start with #-5=log_2x#. Now, we can rewrite a logarithm as an exponentiation equation, like this: #color(red)(y)=color(blue)(b)^(color(green)(x)# becomes #log_color(blue)(b)color(red)(y)=color(green)(x)# and vice versa. So, #log_color(blue)(2)color(red)(x)=color(green)(-5)# becomes #color(red)(x)=color(blue)(2)^(color(green)(-5)#.

And if we solve #2^-5# we get #x=1/32# or #.03125#.

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