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**Unformatted text preview: **write 129 = 2log2 (129) . We’d then have 2x = 2log2 (129) , which means our solution is x = log2 (129).
This makes sense because, after all, the deﬁnition of log2 (129) is ‘the exponent we put on 2 to get
129.’ Indeed we could have obtained this solution directly by rewriting the equation 2x = 129 in
its logarithmic form log2 (129) = x. Either way, in order to get a reasonable decimal approximation
to this number, we’d use the change of base formula, Theorem 6.7, to give us something more
calculator friendly,1 say log2 (129) = ln(129) . Another way to arrive at this answer is as follows
ln(2)
2x = 129
ln (2x ) = ln(129)
x ln(2) = ln(129)
ln(129)
x=
ln(2) Take the natural log of both sides.
Power Rule ‘Taking the natural log’ of both sides is akin to squaring both sides: since f (x) = ln(x) is a function,
as long as two quantities are equal, their natural logs are equal.2 Also note that we treat ln(2) as
any other non-zero real number and divide it through3 to isolate the variable...

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