- #1
cbarker1
Gold Member
MHB
- 349
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Hi Everyone, I need some help using five-place table to find the value of this logarithm. $\log_{10}\left({0.002261}\right)$=
$\log_{10}\left({2.261*10^{-3}}\right)$
I use the multiplication to sum rule for logarithms hence
$(\log_{10}\left({2.261}\right)-3+10)-10$
I look up the value of the log.
$\log_{10}\left({2.261}\right)$+(-3+10)-10=.35430+7-10
However the book said to +10 to the logarithm then subtract 10 from the value; thus, the answer is: 7.35430-10. When I checked on the calculator for the value of the log is -2.64569, i got the wrong answer. Thank for your help.
$\log_{10}\left({2.261*10^{-3}}\right)$
I use the multiplication to sum rule for logarithms hence
$(\log_{10}\left({2.261}\right)-3+10)-10$
N | 0 | 1 | 2 |
220 | 34 242 | 262 | 282 |
221 | 34 439 | 459 | 479 |
222 | 34 635 | 655 | 674 |
223 | 34 830 | 850 | 869 |
224 | 35 025 | 044 | 064 |
225 | 35 218 | 238 | 257 |
226 | 35 411 | 430 | 449 |
$\log_{10}\left({2.261}\right)$+(-3+10)-10=.35430+7-10
However the book said to +10 to the logarithm then subtract 10 from the value; thus, the answer is: 7.35430-10. When I checked on the calculator for the value of the log is -2.64569, i got the wrong answer. Thank for your help.
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