{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Stitz-Zeager_College_Algebra_e-book

# To check this against our previous calculations we nd

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online