Thread: logarithm

1. logarithm

Without a calculator, show that:

log 5 + log 9 + log 8 < 2
log 19 log 19 log 19

2. Originally Posted by math123456
Without a calculator, show that:

log 5 + log 9 + log 8 < 2
log 19 log 19 log 19
$\displaystyle \frac{\log{5} + \log{9} + \log{8}}{\log{19}} = \frac{\log(5 \cdot 9 \cdot 8)}{\log{19}} = \frac{\log{360}}{\log{19}} < \frac{\log{361}}{\log{19}} = \frac{\log{19^2}}{\log{19}} = \frac{2\log{19}}{\log{19}} = 2$

3. Hello, math123456!

Without a calculator, show that: .$\displaystyle \frac{\log5}{\log19} + \frac{\log9}{\log19}+ \frac{\log 8}{\log19} \;<\;2$
Base-Change Formula: .$\displaystyle \log_ba \:=\:\frac{\log a}{\log b}$

The inequality becomes: .$\displaystyle \log_{19}5 + \log_{19}9 + \log_{19}8 \;=\;\log_{19}360$

Since $\displaystyle 19^2 = 361$, then: .$\displaystyle \log_{19}361 \:=\:2$

Then: .$\displaystyle 360 \,<\,361 \quad\Rightarrow\quad \log_{19}360 \,<\,\log_{19}361$

. . Therefore: .$\displaystyle \log_{19}360 \,<\,2$ . . . QED