Question

- log
+ log
-log 7 #
#:​

Answer 1

Answer:

$\displaystyle\log _{10}\left(\frac{11}{5}\right)+\log _{10}\left(\frac{14}{3}\right)-\log _{10}\left(\frac{22}{15}\right)=\log _{10}\left(7\right)\quad \left(\mathrm{Decimal:\quad }\:0.84509\dots \right)$

Step-by-step explanation:

$\displaystyle\log _{10}\left(\frac{11}{5}\right)+\log _{10}\left(\frac{14}{3}\right)-\log _{10}\left(\frac{22}{15}\right)$

$\gray{\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)}$

$\displaystyle\gray{\log _{10}\left(\frac{11}{5}\right)+\log _{10}\left(\frac{14}{3}\right)=\log _{10}\left(\frac{11}{5}\cdot \frac{14}{3}\right)}$

$\displaystyle=\log _{10}\left(\frac{11}{5}\cdot \frac{14}{3}\right)-\log _{10}\left(\frac{22}{15}\right)$

$\displaystyle\gray{\frac{11}{5}\cdot \frac{14}{3}=\frac{154}{15}}$

$\displaystyle=\log _{10}\left(\frac{154}{15}\right)-\log _{10}\left(\frac{22}{15}\right)$

$\displaystyle\gray{\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)}$

$\gray{\log _{10}\left(\dfrac{154}{15}\right)-\log _{10}\left(\dfrac{22}{15}\right)=\log _{10}\left(\dfrac{\dfrac{154}{15}}{\dfrac{22}{15}}\right)}$

$=\log _{10}\left(\dfrac{\dfrac{154}{15}}{\dfrac{22}{15}}\right)$

$\gray{\dfrac{\dfrac{154}{15}}{\dfrac{22}{15}}=7}$

$=\log _{10}\left(7\right)$