SOLUTION: Write each expression in terms of common logarithms, and then give a calculator approximation (correct to four decimal places).
(a)log8(10)
exact value
(b)log6 (302)
exact valu
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Question 890808: Write each expression in terms of common logarithms, and then give a calculator approximation (correct to four decimal places).
(a)log8(10)
exact value
(b)log6 (302)
exact value
Found 2 solutions by MathLover1, Theo:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
(a)
=
=
exact value rounded to four decimal places =>
(b)
=
=
exact value rounded to four decimal places =>
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
LOG = log to the base of 10 that your calculator can handle.
LN = log to the base of e that your calcjulator can handle.
most scientific calculators can handle both of these.
convert to one or the other and you can find the log using your calculator.
log8(10) = LOG(10) / LOG(8)
log8(10) = LN(10) / LN(8)
both will get you the same answer.
log6(302) = LOG(302) / LOG(6)
log6(302) = LN(302) / LN(6)
both will get you the same answer.
LOG(10)/LOG(8) = LN(10)/LN(8) = 1.107309365 = 1.1073 rounded to 4 decimal places.
LOG(302)/LOG(6) = LN(302)/LN(6) = 3.187050001 = 3.1871 rounded to 4 decimal places.
those answers are exact to the best ability of the calculator.
they could contain more decimal places than shown on the display.
they do not appear to be rational although i can't tell for sure.
the calculator couldn't give me an equivalent expression in terms of fractions so i'm assuming they are irrational and can't be displayed as a ratio of 2 integers.
in exponential form, the 2 expressions would look like this:
log8(10) = y if and only if 8^y = 10
log6(302) = y if and only if 6^y = 302
to solve for y, you would have to convert them back to logs again.
you will wind up with the conversion formula in both cases.
for example:
8^y = 10
take log of both sides to get:
LOG(8^y) = LOG(10)
since LOG(8^y) = y*LOG(8), equation becomes:
y*LOG(8) = LOG(10)
divide both sides of equation by LOG(8) to get:
y = LOG(10)/LOG(8)
that's the same as using the log of any base to LOG conversion formula.
that formula is:
logb(x) = LOG(x)/LOG(b)
it's the same formula we used up top.
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