Logarithms

1) Find the value of log 6 (1296 * 46656)

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  log  6 (1296 * 4665 6 ) = log  6 1296 + log  6 4665 6  

    = log  6 6 4 + log  6 6 6
   = 4 log  6 6 + 6 log  6 6 = 10 .  Hence (a).

2) \({1 \over {3}}\) log10 8 + log10 15 - log10 3 equals to

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\({1 \over {3}}\) log 10 8 + log 10 15 - log 10 3  = log 10 8 (1/3) + log 10 15 - log 10 3

= log 10 (2*15) - log 10 3

= log 10   \({2*15 \over {3}}\) =  log 10 10 = 1  Hence (b).

3) log (\({p{}{}{}{}{}{}{}{}{}{}{}{} \over {qr}}\)) + log (\({q \over {pr}}\)) + log (\({r \over {pq}}\)) – log (pqr) is

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log ( \({p{}{}{}{}{}{}{}{}{}{}{}{} \over {qr}}\) ) + log ( \({q \over {pr}}\) ) + log ( \({r \over {pq}}\) ) - log (pqr)  = log ( \({p \over {qr}}\) *  \({q \over {pr}}\)   *  \({r \over {pq}}\) ) - log (pqr) = 0

Hence (d).

4) If (log4 a)( loga 3a)( log3a b) = loga a3, then b equals

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  (log 4 a )(   log a 3a)(   log 3a b) = log a   a 3

=>  \({loga \over {log4}}\) *  \({log3a \over {loga}}\) *  \({logb \over {log3a}}\)   =  \({loga \over {a}}\)

  =>  \({logb \over {log4}}\) = 3

=>  log b = 3 log 4

= log 4 3  

b = 64 .  Hence (c).

 

5) If log2 = 0.301 and log3 = 0.477, then find 3log36.

Show Solution
  3log36 = 3log 10   (4*9 )

= 3log 10 4 + 3log 10   9

= 3 * 2 log 10 2 + 3 * 2  log 10 3

= (6 * 0.301) + (6 * 0.477)

= 1.806 + 2.862

= 4.668.  He nce (b).

6) If x = loga (\({4b \over {a}}\)) + loga (\({a \over {b}}\))

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  x = log a ( \({4b \over {a}}\) ) + log a ( \({a \over {b}}\) )
    = log a ( \({4b \over {a}}\) * \({a \over {b}}\) ) = log a 4ab .  Hence (c).

7)If log (x-8) + log (x+6) = log (6x+17), then find the value of x.

Show Solution
    log 10 (x-8) + log 10 (x+6 ) = log 10 (6x+17) 
  ( x-8 )(x+6) = 6x + 17
    X 2 + 6x – 8x – 48 = 6x + 17
  X 2 - 8 x – 65 = 0
  ( x-13 )(x+5)= 0
X = 13, -5 .   Hence (b).

8)

If \({loga \over {6}}\) = \({logb \over {4}}\) = \({logc \over {8}}\), then bc in terms of a is

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\({loga \over {6}}\) = \({logb \over {4}}\) = \({logc \over {8}}\) = k => log a = 6k, log b = 4k, log c = 8k => log bc = 4k + 8k = 12k => log a2 = 12k = log bc à bc = a2 Hence (c).

9) If log7 y + log7 \({1 \over {8}}\) = 1, then the value of y is

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log 7 y + log 7   \({1 \over {8}}\) = 1

\({logy \over {log7}}\) +  \({log({1 \over {8}}) \over {log7}}\) = 1

\({logy+log1-log8 \over {log7}}\) = 1

log y + log 1 – log 8 = log 7
  • log y + log 1 = log 7 + log 8


  • = log 56
    l og y = log 56 – log 1 = log 56
    y = 56 .  Hence (d).

    10) If log36 6, log216 6 and loga 6 are in HP respectively, find the value of a.

    Show Solution
    If x, y and z are in HP we can write it as y =  \({2xz \over {x+z}}\)

    \({2xz \over {y}}\) = x+z =>  \({2 \over {y}}\) =  \({x+z \over {xz}}\)  

     \({2 \over {y}}\) =  \({1 \over {x}}\) +  \({1 \over {z}}\)

    \({2 \over {log_{216} 6}}\) =  \({1 \over {log_{36} 6}}\) +  \({1 \over {log_{a} 6}}\)  

    2log 6 216 = log 6 36 * log 6 a

    216 2 = 36a

    a = 1296 .  Hence (a).

    11)

    What is the value of m, if log4 16 * log16 64 * log64 256 *….. up to mth term = 53?

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    log4 16 * log16 64 * log64 256 *….. up to mth term = 53 \({log16 \over {log4}}\) * \({log64 \over {log16}}\) * \({log256 \over {log64}}\) ….\({log4{}{}{}{}{}{}{}{}{}{}{}{} \over {log4}}\) = 53 \({log4 \over {log4}}\) = 53 \({(m+1)log4 \over {log4}}\) = 53 à m = 52. Hence (d).

    12)

    Solve for y: log16(2 log8(7+ log4(3+\({1 \over {3}}\)(log5y)))) = \({1 \over {4}}\)

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    log16(2 log8(7+ log4(3+\({1 \over {3}}\)(log5y)))) = \({1 \over {2}}\) (2 log8(7+ log4(3+\({1 \over {3}}\)(log5y)))) = 16(1/4) = 2 log8(7+ log4(3+\({1 \over {3}}\)(log5y))) = \({2 \over {2}}\) = 1 7+ log4(3+\({1 \over {3}}\)(log5y))= 81 = 8 log4(3+\({1 \over {3}}\)(log5y)) = 8 – 7 = 1 3+\({1 \over {3}}\)(log5y) = 41 = 4 \({1 \over {3}}\)(log5y) = 4-3 = 1 log5y(1/3) = 1 y(1/3) = 5 à y = 53 = 125. Hence (c).

    13) What is the value of p, if \({1 \over {log_{({345 \over {343}})}p}}\) + \({1 \over {log_{({347 \over {345}})}p}}\) + \({1 \over {log_{({349 \over {347}})}p}}\) + …..+ \({1 \over {log_{({729 \over {727}})}p}}\) = 3?

    Show Solution
    \({1 \over {log_{({345 \over {343}})}p}}\) =  \({log_{343}^{345} \over {logp}}\) = log p   \({345 \over {343}}\)

    log p   \({345 \over {343}}\) + log p   \({347 \over {345}}\) + log p   \({349 \over {347}}\) +… log p   \({729 \over {727}}\) = 3

    log p (   \({345 \over {343}}\) *  \({347 \over {345}}\) *  \({349 \over {347}}\) *…*  \({729 \over {727}}\) = 3

    log p   \({729 \over {343}}\) = 3

    p 3 =  \({729 \over {343}}\) = p  =  \({9 \over {7}}\) .  Hence (a).

    14)

    If log2064 = x, then log5100 is

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    log2064 = x log2043 = x

    log204 = \({x \over {3}}\)

    log420 =\({3 \over {x}}\)

    log45 + log44 =\({3 \over {x}}\)

    log45 =\({3-x \over {x}}\) ……………….(i)

    Let X = log5100 log520 + log55 = X

    log520 = X – 1

    log54 + log55 = X – 1

    log54 = X – 2

    log45 = \({1 \over {X-2}}\) ………………(ii)

    Equate (i) and (ii) \({3-x \over {x}}\) = \({1 \over {X-2}}\) X = \({6-x \over {3-x}}\). Hence (b).

    15)

    If log102 = S and log103 = P, then find the value of log536.

    Show Solution

    log536 = log5(4*9) = log54 + log59 = log522 + log532 = 2log52 + 2log53

    \({2log_{10} 3 \over {log_{10} 5}}\) + \({2log_{10} 2 \over {log_{10} 5}}\)

    \({2log_{10} 3 \over {log_{10} 10-log_{10} 2}}\) + \({2log_{10} 2 \over {log_{10} 10-log_{10} 2}}\)

    \({2P \over {1-S}}\) + \({2S \over {1-S}}\)

    \({2P+ 2S \over {1-S}}\) = \({2(P+S) \over {1-S}}\). Hence (d).