#### 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.

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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.

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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.

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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).

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