Question

write 2 log 3+3 log 5-5log2 as a single logarithm.
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Answer 1

Answer:

$2\log _{10}\left(3\right)+3\log _{10}\left(5\right)-5\log _{10}\left(2\right)=\log _{10}\left(\dfrac{1125}{32}\right)\quad \left(\mathrm{Decimal:\quad }\:1.54600\dots \right)$

Step-by-step explanation:

$2\log _{10}\left(3\right)+3\log _{10}\left(5\right)-5\log _{10}\left(2\right)$

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

$\gray{2\log _{10}\left(3\right)=\log _{10}\left(3^2\right)}$

$=\log _{10}\left(3^2\right)+3\log _{10}\left(5\right)-5\log _{10}\left(2\right)$

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

$\gray{3\log _{10}\left(5\right)=\log _{10}\left(5^3\right)}$

$=\log _{10}\left(3^2\right)+\log _{10}\left(5^3\right)-5\log _{10}\left(2\right)$

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

$\gray{\log _{10}\left(3^2\right)+\log _{10}\left(5^3\right)=\log _{10}\left(5^3\cdot \:3^2\right)}$

$=\log _{10}\left(5^3\cdot \:3^2\right)-5\log _{10}\left(2\right)$

$\gray{3^2\cdot \:5^3=1125}$

$=\log _{10}\left(1125\right)-5\log _{10}\left(2\right)$

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

$\gray{5\log _{10}\left(2\right)=\log _{10}\left(2^5\right)}$

$=\log _{10}\left(1125\right)-\log _{10}\left(2^5\right)$

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

$\gray{\log _{10}\left(1125\right)-\log _{10}\left(2^5\right)=\log _{10}\left(\dfrac{1125}{2^5}\right)}$

$=\log _{10}\left(\dfrac{1125}{2^5}\right)$

$\gray{2^5=32}$

$=\log _{10}\left(\dfrac{1125}{32}\right)$

Answer 2

Step-by-step explanation:

✍️log10(1125/32) as a single logarithm....

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