Question 969711: May I know how to answer this question?
Given that log12(3)=p,express each of the following in terms of p
(i)log sqrt12 (1/9)
(ii)log1/12(27)
(iii)log12(1/4)
(iv) log144(4)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you are given that:
log.12(3) = p
you are asked to find the following:
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number 1:
find log.sqrt(12)(1/9) in terms of p.
log.sqrt(12)(1/9) = y if and only if sqrt(12)^y = 1/9
sqrt(12)^y is the same as 12^(y/2), so the equation becomes:
12^(y/2) = 1/9
square both sides of this equation to get 12^y = 1/81.
this is true if and only if log.12(1/81) = y
1/81 is the same as 3^-4, so this equation becomes:
log.12(3^-4) = y
this is the same as -4*log.12(3) = y
since log.12(3) = p, this is the same at -4*p = y
since y = -4*p, then log.sqrt(12)(1/9) = y becomes log.sqrt(12)(1/9) = -4*p.
that's your solution to the first problem.
log.sqrt(12)(1/9) = -4*p.
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problem number 2:
find log.1/12(27) in terms of p.
set this equal to y and you get log.1/12(27) = y
this is true if and only if (1/12)^y = 27
1/12 is the same as 12^-1, so the equation becomes:
12^-1)^y = 27
this is the same as 12^-y = 27
this is the same as 1/12^y = 27
solve for 12^y to get 12^y = 1/27
this is true if and only if log.12(1/27) = y
1/27 is the same as 3^-3, so the equation becomes:
log.12(3^-3) = y
this is the same as -3*log.12(3) = y
log.12(3) = p, so this equation becomes -3*p = y
since y = -3*p, then the equation of log.1/12(27) = y becomes log.1/12(27) = -3*p.
that's your solution.
log.1/12(27) = -3*p
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problem number 3:
find log.12(1/4) in terms of p.
set 3^x = 1/4 and solve for x.
take log of both sides of this equation to get log(3^x) = log(1/4)
this is the same as x * log(3) = log(1/4).
solve for x to get x = log(1/4)/log(3).
your equation of log.12(1/4) is equal to log.12(3^(log(1/4)/log(3)).
this is the same as log.12(1/4) is equal to log(1/4)/log(3) * log.12(3).
since log.12(3) is equal to p, then this equation becomes:
log.12(1/4) is equal to log(1/4)/log(3) * p.
that's your solution.
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problem 4:
find log.144(4) in terms of p.
log.144(4) = y if and only if 144^y = 4
this is the same as log.144(4) = y if and only if (12^2)^y = 4 which is the same as 12^(2y) = 4.
this is true if and only if log.12(4) = 2y
divide both sides of this equation by 2 to get 1/2 * log.12(4) = y
this is the same as log.12(4^(1/2) = y which is the same as log.12(2) = y
set 3^x = 2 and solve for x.
take the log of both sides of this equation to get log(3^x) = log(2).
this is the same as x*log(3) = log(2).
solve for x to get x = log(2)/log(3).
your equation of log.12(2) = y becomes log.12(3^(log(2)/log(3)) = y.
log.12(3^(log(2)/log(3)) = y is the same as log(2)/log(3)*log.12(3) = y.
since log.12(3) is equal to p, then log(2)/log(3)*log.12(3) = y becomes log(2)/log(3)*p = y.
since y = log(2)/log(3)*p and since we started with log.144(4) = y, then we get:
log.144(4) = log(2)/log(3)*p.
that's your solution.
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to summarize your solutions.
you are given that log.12(3) = p
you are asked to find the following in terms of p.
(i)log sqrt12 (1/9)
(ii)log1/12(27)
(iii)log12(1/4)
(iv) log144(4)
your solutions are:
(i)log sqrt12 (1/9) = -4*p
(ii)log1/12(27) = -3*p
(iii)log12(1/4) = log(1/4)/log(3) * p
(iv) log144(4) = log(2)/log(3)*p
you can confirm all of these using your calculator and the log conversion formula of log.b(x) = log(x)/log(b), where log is the LOG function of your calculator.
p = log.12(3) which is equal to log(3)/log(12) which is equal to .4421141087
log.sqrt(12)(1/9) = log(1/9)/log(sqrt(12)) = -1.768456435
-4*p = -4*.4421141087 = -1.768456435
they match.
log.1/12(27) = log(27) / log(1/12) = -1.326342326
-3*p = -3*.4421141087 = -1.326342326
they match.
log.12(1/4) = log(1/4) / log(12) = -.5578858913
log(1/4) /log(3) * p = -.5578858913
they match.
log.144(4) = log(4)/log(144) = .2789429457
log(2)/log(3)*p = .2789429457
they match
all the answers match between the original equations and the final equations so the solutions are good.
in case you haven't figured it out yet, log.b means log to the base of b.
similarly log.12 means log to the base of 12.
similarly log.sqrt(12) means log to the base of sqrt(12).
similarly log.1/12 means log to the base of 1/12.
etc.
i use the . to separate the base from log.
it makes it a little easiwer to read.
log1/12 would have been cumbersom.
log.1/12 makes it easier to see that the base is 1/12.
this laso avoids the use of log[1/12](x) which is technically more correct but also more difficult to type.
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