Evaluate the given expression: [tex](log_2\frac{1}{3})(log_3\frac{1}{4})(log_4\frac{1}{5})(log_5\frac{1}{6})(log_6\frac{1}{7})(log_7\frac{1}{8})[/tex].
Please answer with solution, thanks in advance! :))
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Evaluate the given expression: [tex](log_2\frac{1}{3})(log_3\frac{1}{4})(log_4\frac{1}{5})(log_5\frac{1}{6})(log_6\frac{1}{7})(log_7\frac{1}{8})[/tex].
Please answer with solution, thanks in advance! :))
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Answer:
We can write all in terms of "log".
[tex]\sf log _2\left(\frac{1}{3}\right)\to - log _2\left(3\right)[/tex]
Similarly, write all in terms of log and then multiplying those "-" signs, we get:
[tex]\sf log_23 \cdot log_34 \cdot log_45 \cdot log_56 \cdot log_67 \cdot log_78[/tex]
Convert them into fractions.
[tex]\sf \dfrac{log3}{log2} \cdot \dfrac{log4}{log3} \cdot \dfrac{log5}{log4} \cdot \dfrac{log6}{log5} \cdot \dfrac{log7}{log6} \cdot \dfrac{log8}{log7}[/tex]
Many terms eventually cancel out! Simplify this equation, you will get:
[tex]\sf \dfrac{log8}{log2}[/tex]
[tex]\sf log _2(8)[/tex]
[tex]\sf 3[/tex]
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