How do you solve # log_2(24) - log_2(3) = log_2x#?

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Answer 1 · 2015-12-03T13:41:24

#x = 8#

Use the logarithmic law

#log_a(x) - log_a(y) = log_a(x/y)#

Thus,

#log_2(24) - log_2(3) = log_2(x)#

#<=> log_2(24/3) = log_2(x)#

#<=> log_2(8) = log_2(x)#

At this point, you already see that #x = 8#.

To make it even more clear, let's get rid of the logarithmic expressions.

The inverse function of #log_2(x)# is #2^x# which means that #log_2(2^x) = x# and also #2^(log_2(x)) = x# hold.

This means:

#log_2(8) = log_2(x)#

#<=> 2^(log_2(8)) = 2^(log_2(x))#

#<=> 8 = x#