Without-using-a-calculator-evaluate-log-5-2-log-2-log-50-

Question Number 5579 by Rasheed Soomro last updated on 21/May/16

Without using a calculator, evaluate

{(\log 5)}^{2} + \log 2 \log 50

Answered by FilupSmith last updated on 21/May/16

\log x = \log_{10} x

\log {(5)}^{2} + \log (5) \log (50)

= \log (5) (\log (5) + \log (50))

= \log (5) \log (250)

\log (250) = \log (100 \times 2.5)

= \log (100) + \log (2.5)

= 2 + \log (2.5)

∴ \log (5) \log (250) = \log (5) (2 + \log (2.5))

= 2 \log (5) + \log (5) \log (2.5)

c o n t i n u e

Commented by nchejane last updated on 21/May/16

I t i s l i k e i t i s n o t c o p i e d c o r r e c t l y

{(l o g 5)}^{2} + l o g 2 l o g 50

= {(l o g 5)}^{2} + l o g 2 (l o g (5 \times 10))

= {(l o g 5)}^{2} + l o g 2 (l o g 5 + l o g 10)

= {(l o g 5)}^{2} + l o g 2 l o g 5 + l o g 2

= (l o g 5) [l o g 5 + l o g 2] + l o g 2

= l o g 5 l o g 10 + l o g 2

= l o g 5 + l o g 2

= l o g 10

= 1

Commented by Rasheed Soomro last updated on 21/May/16

G^{\overset{v}{OO}} D!