Mathematics Must do problems of logarithm for NDA

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Must do problems of logarithm for NDA

Q 2815167969

Consider the equation `x^(log_5 x^2 + ( log_5 x )^2 -12) = 1/x^4`
The number of solutions of the equation is

(A)

(B)

(C)

(D)

Solution:

We have

`x^(log_5 x^2 + ( log_5 x )^2 -12) = 1/x^4`

Taking log to the base 5 of both the sides, we get

`[log_5 x^ 2 + (log_5 x)^2- 12]log_5 x`

`= -4log_5x`

`=> [log_5x^2 + (log_5x)^2 - 8 ] log_5 x`

`=0`

`=> log_5 x = 0`

or `(log_5 x)^2 + 2 log _5 x-8 = 0`

`=> x = 5^0`

or ` ( log_5 x -2) ( log_5 x + 4 ) =0`

`=> x =1 ` or `log _5x =2 , -4`

`=> x =1 , 25 , 1/625`

Number of solutions are 3.

Correct Answer is `=>` (C) 3

Q 2815167969

Consider the equation `x^(log_5 x^2 + ( log_5 x )^2 -12) = 1/x^4`
The set of all x satisfying the equation is

(A)

`{1,25, 1/125, 1/625}`

(B)

`{1,25,1/625}`

(C)

`{1, 25}`

(D)

`{1}`

Correct Answer is `=>` (B) `{1,25,1/625}`

Q 2815167969

Consider the equation `x^(log_5 x^2 + ( log_5 x )^2 -12) = 1/x^4`
The product of all values of x is

(A)

(B)

(C)

1/25

(D)

1/3125

Correct Answer is `=>` (C) 1/25

Q 2875167966

Let `a, b, c in R; a != b != c` and `(logx)/(b -c) = (log y)/ (c -a) = ( log z)/(a-b)`
The value of `x^a · y^b · z^c` is

(A)

(B)

(C)

(D)

None of these

Solution:

Let `(log x)/(b -c) = (log y)/(c -a) = (log z)/(a -b) = k`

`=> log x = k(b - c)=> x = 10^(k(b-c))`

`log y = k(c -a) => y = 10^(k(c-a))`

`log z = k ( a -b) => z = 10^(k(a -b))`

Now, `x^a * y^b * z^c`

`= 10^(k(b -c)a + k(c -a) b + k(a -b) c)`

`= 10^(k[a(b -c) + b(c -a) + c (a -b) ])`

`= 10^0 = 1`

Correct Answer is `=>` (B) 1

Q 2875167966

Let `a, b, c in R; a != b != c` and `(logx)/(b -c) = (log y)/ (c -a) = ( log z)/(a-b)`
The value of `1/(log_(xy) 2) + 1/ (log_(yz) 2 ) + 1/ (log_(zx) 2 )`

(A)

(B)

(C)

(D)

None of these

Solution:

`1/(log_(xy) 2) + 1/ (log_(yz) 2 ) + 1/ (log_(zx) 2 )`

`= log_2 xy + log_2 yz + log_2 zx`

`= log_2 (xy xx yz xx zx)`

`= log_2 (xyz)^2 = 2log_2 (xyz)`

`= 2 log _2 1 =0 \ \ \ \ \ \ [ because xyz =1 ]`

Correct Answer is `=>` (A) 0

Q 2815167960

If `x in [- 2pi, 2pi ]` and `log_0.5 sin x = 1 - log_0.5 cos x`, then consider the following statements

I. Number of solutions in the interval `[- 2pi, 2pi]` is 1.

II. One solution of the equation is `pi//4`.

Which of the above statement(s) is/are correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

Clearly, `log_(0.5) sin x` and `log_(0.5) cos x` are defined, if `x in (0, pi//2)`

Now, `log_(0.5) sin x = 1 - log_(0.5) cos x`

`=> log_(0.5) sinx +log_(0.5) cos x = 1`

`=> log_(0.5) ( sin x xx cos x ) = 1`

`=> sin x cos x = ( 0.5)^1`

`=> sin 2x = 1`

`=> x = pi/4, (3 pi)/4`

Hence, only Statement II is correct.

Correct Answer is `=>` (B) Only II

Q 2845067863

If `log_a (n -k) < log_(a^ 2) (n - k)` and `a > 1`, then n lies in the interval

(A)

`[k, k+1]`

(B)

`(k, k -1)`

(C)

`(k, k +1)`

(D)

None of these

Solution:

`log_a (n- k) < log_(a^ 2) (n- k)`

`=> ( n - k)^2 < n - k \ \ \ \ \ \ [ because a > 1 ]`

`=> n^2 - 2 kn + k^2 < n -k`

`=> n^2 - kn - ( k +1) n + k ( k +1) < 0`

`=> ( n -k ) ( n - k -1) < 0`

`=> k < n < k +1`

So, n lies in (k, k + 1).

Correct Answer is `=>` (C) `(k, k +1)`

Q 2815867760

If `log_(1//3) [log_4 (x^2- 12)] > 0`, then x lies m the interval

(A)

`(-oo, -4) uu ( 4, oo)`

(B)

`( -oo, oo)`

(C)

`(- oo, -2) uu ( 2, oo)`

(D)

None of these

Solution:

Given expression is

`log_(1//3) [ log_4 ( x^2 - 12) ] > 0`

`because x^2 - 12 > 4`

`=> x^2 - 16 > 0`

`=> | x | > 4`

`=> x < -4 ` or ` x > 4`

So, x lies in `(-oo , -4) uu ( 4, oo)`

Correct Answer is `=>` (A) `(-oo, -4) uu ( 4, oo)`

Q 2885667567

If `x = log_3 5, y = log_17 25`, then which one of the following is correct'?

(A)

x < y

(B)

`x = y`

(C)

x > y

(D)

None of these

Solution:

Given, `y = 2log_17 5` and `x = log_3 5`

`:. 1/y =1/2 log_5 17`

and `1/x = log_5 3 = 1/2 log_5 9`

Clearly `1/y > 1/x`

`:. x > y`

Correct Answer is `=>` (C) x > y

Q 2825667561

If `x^18 = y^21 = z^28` , then `3 3 log y^x, 3log_z y, 7 log_x z` are in

(A)

(B)

(C)

(D)

None of these

Solution:

Let `x^18 =y^21 = z^28 = k` , then

`18 logx = 21 logy = 28 logz = log k`

`=> log x = (log x)/18 , log y = (log k ) /21`

and `log z = (log k)/28`

`:. 3 log _y x = ( 3 log x)/ ( log y) = ( 3 xx 21 )/18 = 7/2`

`3 log_z y = ( 3 log y ) / ( log z ) = ( 3 xx 28 ) /21 =4`

`7 log _x z = ( 7 logz)/ ( logx ) = ( 7 xx 18 ) /28 = 9/2`

So `3, 7/2, 4 , 9/2` are in A.P

Correct Answer is `=>` (A) AP

Q 2815467369

`(log x + log x^4 + log x^9 + .... + logx^(n^2))/(log x + logx^2 + logx^3 + .... + log x^n) ` is equal to

(A)

`(2n+1)/3`

(B)

`(2n -1)/3`

(C)

`(3(n +2))/2`

(D)

`(3(n -1))/2`

Solution:

Given expression

`= ( (1+ 4+9 +... + n^2) log x)/(( 1+ 2 +3 + ... + ) logx )`

`=( Sigman^2)/(Sigman) = (( n( n+1 ) ( 2 n + 1))/6)/((n(n+1))/2)`

`= (2n +1)/3`

Correct Answer is `=>` (A) `(2n+1)/3`

Q 2845367263

If `1/(log_a x) + 1/( log_c x) = 2/(log_b x)`. then a ,b and c are in

(A)

(B)

(C)

(D)

None of these

Solution:

`log_x a + log_x c = 2 log_x b`

`=> ac =b^2`

i.e a, b and c are in G.P

Correct Answer is `=>` (B) GP

Q 2885267167

If `log_12 27 =a`, then `log_6 16` is equal to

(A)

` 2* (3-a)/(3+a)`

(B)

`3 * ( 3 -a)/( 3 +a)`

(C)

`4 * ( 3 -a) / (3 +a) `

(D)

None of these

Solution:

`log_6 16 = log _6 2^4 = 4 log_6 2 = 4/(log_2 6)`

`= 4/(log_2 2 + log_2 3) = 4/ ( 1+ log_2 3)` ........(i)

and ` a = log_12 27 = log_12 3^3 = 3 log_12 3`

`= 3/(log_3 12) = 3/ ( log_3 3 + log_3 4)`

`= 3/ (1 + 2 log_3 2)`

`=> a + 2a log_3 2 =3`

`=> log_3 2 = (3 -a)/ (2a)`

`:. log_2 3 = ( 2a) / ( 3 -a)`

from Eq (i)

`log_6 16 = 4/ ( 1 + (2a)/( 3-a))`

`= (4(3 -a) )/ ( 3 +a)`

Correct Answer is `=>` (C) `4 * ( 3 -a) / (3 +a) `

Q 2885167067

`log_10 tan1°+ log_10 tan 2° + ... + log_10 tan 89°` is equal to

(A)

(B)

(C)

(D)

Solution:

`log tan 89° = logcot 1° = -log tan 1°`

`:.` Given expression becomes

`log tan 1° + log tan 2° + ... + log tan 44 ° + log tan 45° -`` log tan 44° - ..... - log tan 2° - log tan 1°`

`= log tan 45° = log 1 = 0`

Correct Answer is `=>` (A) 0

Q 2815167060

If `a = log_24 12, b = log_36 24, c = log_48 36`. Then `1 + abc` is equal to

(A)

2 ac

(B)

2 bc

(C)

2 ab

(D)

None of these

Solution:

`abc = ( log 12 ) /( log 24) * ( log 24 )/ ( log 36) * ( log 36)/ ( log 48 ) = (log 12)/ ( log 48)`

`:. 1 + abc = ( log 48 + log 12)/ ( log 48)`

`= (log ( 48 * 12))/ ( log 48 ) = (log 24^2)/(log 48)`

`= 2 * ( log 24)/ ( log 48 ) = ( log 24) / ( log 36 ) xx (log 36)/ ( log 48 )`

`= 2 log_36 24 xx log _48 36 = 2 bc`

Correct Answer is `=>` (B) 2 bc

Q 2865056865

The value of `81 ^(1//log_5 3) + 27^(log_9 36) + 3 ^(4// log_7 9)` 9 is equal to

(A)

(B)

625

(C)

216

(D)

890

Solution:

Let `T_1 = 3^(4log_3 5) = 3^(log_3 (5)^4) =5^4 = 625`

`T_2 = (3^3)^(2/2 log_3 6 = 3 log_3 6^3 = 216`

`T_3 = 3^4 log_9 7 = 3^(4 * 1/2 * log_3 7)`

`= 3 log_3 7^2 = 49`

`:. T_1 + T_2 + T_3 = 625 + 216 + 49 = 890`

Correct Answer is `=>` (D) 890

Q 2865856765

The value of `(yz )^(log y - log z) xx (zx)^(log z - log x) xx (xy)^(log x - log y) ` is equal to

(A)

(B)

(C)

(D)

Correct Answer is `=>` (B) 1

Q 2825856761

`(log_8 17)/(log_9 23) - (log_(2sqrt2) 17)/( log_3 23)` is equal to

(A)

(B)

(C)

`17/8`

(D)

`23/17`

Solution:

`(1/3 log_2 17 )/( 1/2 log_3 23) - ( 2/3 log_2 17 )/ ( log _3 23) = 0`

`=> (2/3 log_2 17 )/ ( log_3 23) - ( 2/3 log_2 17 ) / ( log_3 23) = 0`

Correct Answer is `=>` (A) 0

Q 2805656568

If `log_10 2 = 0.30103`, then `log_10 50` is equal to

(A)

2.30103

(B)

2.69897

(C)

1.69897

(D)

0.69897

Solution:

`log_10 50 = log_10\ \ 100/2`

`= log _ 10 100 - log _10 2`

`= 1.69897`

Correct Answer is `=>` (C) 1.69897

Q 2563080845

If `alpha , beta` are the roots of the equation
`ax^2 + bx + c = 0`, then `log (a-bx + cx^2)` is equal to
UPSEE 2010

(A)

`log a+ (alpha + beta) x + (alpha^2 + beta^2 )/2 x^2 + (alpha^3 + beta^3)/3 x^3 +....`

(B)

`log a +(alpha + beta) x - ((alpha^2+ beta^2 )/2) * x^2 + ((alpha^3 + beta^3)/3) x^3 - ......`

(C)

`log a - (alpha + beta ) x- ((alpha^2 + beta^2)/2) x^2 - ((alpha^3 + beta^3)/3) x^3 - ....`

(D)

none of the above

Solution:

Since, `alpha , beta` are roots of the equation

`ax^2 + bx + c = 0`, we have

`:. alpha + beta = (-b)/a , alpha beta =c/a`

`:. a- bx +cx^2 =a (1- b/a x + c/a x^2)`

`=a {1+ (alpha + beta) x + alpha beta x^2 } = a { (1+alpha x)(1+ beta x) }`

Hence, `log (a-bx + cx^2)`

`= log {a (1+alpha x)(1+ beta x) }`

`= log a + log (1+ ax) + log (1+ beta x)`

`= log a + (alpha x - (alpha x)^2/2 + (alpha x)^3/3 -..........)`

`+ (beta x - (beta x)^2/2 + (beta x)^3/3 -............)`

`= log a + (alpha + beta) x - ((alpha^2 + beta^2)/2) x^2 + ((alpha^3 + beta^3)/3) x^3 -.....`

Correct Answer is `=>` (B) `log a +(alpha + beta) x - ((alpha^2+ beta^2 )/2) * x^2 + ((alpha^3 + beta^3)/3) x^3 - ......`

Q 2513601540

The sequence `log a, log a^2/b , log a^3/b^2,.......` is
WBJEE 2011

(A)

a GP

(B)

an AP

(C)

a HP

(D)

both a GP and a HP

Solution:

Let `S= log a, log a^2/b, log a^3/b^2`,.....

`= log a, (2 log a - log b)`,

`(3 log a -2 log b),......`

Now, `T_2 -T_1 = log a - log b`

and `T_3 -T_2 = log a - log b`

Hence, it is an AP.

Correct Answer is `=>` (B) an AP

Q 2572134936

The number of digits in `20^(301)` (given, `log_10 2 =0.3010` ) is
WBJEE 2014

(A)

`602`

(B)

`301`

(C)

`392`

(D)

`391`

Solution:

Let `y = 20^301`

Number of digits= Integral part of `(301 log_10 20) + 1`

=Integral part of `[301 (log_10 10 + log_10 2)] + 1`

=Integral part of `[301(1 + 0.3010)] + 1`

=Integral part of `[301 xx 1.301 0]+ 1`

=Integral part of `[391.601] + 1`

`=391 + 1 =392`

Correct Answer is `=>` (C) `392`

Q 2419191910

If `x= log_a bc, y = log_b ca ` and `z =log_c ab`, then value of `1/(1+x) + 1/(1 +y) + 1/(1 +z)` will be
BCECE Stage 1 2014

(A)

x + y + z

(B)

(C)

`ab + bc + ca`

(D)

`abc`

Solution:

Here, `1 + x =log_a a + log_a bc = log_a abc`

`=> 1/(1 +x) = log_(abc) a`

Similarly, `1/(1 +y) = log_(abc) b`

and `1/(1 +z) = log_(abc) c`

`:. 1/(1 +x) + 1/(1 +y) + 1/(1 +z)`

`= log_(abc) a + log_(abc) b +log _(abc) c`

`= log_(abc) abc = 1`

Correct Answer is `=>` (B) 1

Q 2414523459

The value of `4 + 2(1 + 2) log 2+ (2(1+2^2))/(2!) (log 2)^2+(2(1+2)^3)/(3!)(log 2)^3..........+` is
UPSEE 2011

(A)

`10`

(B)

`12`

(C)

`log (3^2 * 4^2)`

(D)

`log (2^2 *3^2)`

Solution:

`4+ 2 (1+2) log 2+(2 (1+2^2) (log 2)^2)/(2 !)+(2 (1+2^3)(log 2)^3)/(3!)+........`

`=2(1+ log 2 +((log 2)^2)/(2!)+..........)+ 2(1+2 log 2 +((2 log 2)^2)/(2!) +............)`

`2*(e^(log 2)) + 2 (e^(2log 2))`

`= 2 xx 2 + 2e^(log 4)`

`= 4+ 2 xx 4= 12`

Correct Answer is `=>` (B) `12`

Q 2500778618

The number of real roots of equation

`log_(e) x + ex = 0` is
WBJEE 2015

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`3`

Solution:

We have, `log_e x + ex = 0`

`=> log_e x = -ex`

Since, both the graphs intersect at only one point.

Hence, the number of real roots of equation is one.

Correct Answer is `=>` (B) `1`

Q 2550778614

If `log_(0.2) (x-1) > log_(0.04) (x+5)`, then
WBJEE 2015

(A)

`-1 < x < 4`

(B)

`2 < x < 3`

(C)

`1 < x < 4`

(D)

`1 < x < 3`

Solution:

We have, `log_(0.2) (x-1) > log_(0.04) (x+5)`

`=> log_(0.2) (x-1) > log_(0.2)^2 (x+5)`

`=> log_(0.2 ) (x-1) > 1/2 log_(0.2) (x+5)`

`=> 2 log_(0.2) (x-1) > log_(0.2) (x+5)`

`=> log_(0.2) (x-1)^2 > log_(0.2) (x+5)`

`=> (x-1)^2 < x +5`

`[ :. log_(a) x > log_(a) y => x < y , text(if) 0 < a < 1 ]`

`=> x^2 -2x +1 < x+5`

`=> x^2 -3x -4 < 0`

`=> x^2 -4x +x -4 < 0`

`=> x (x-4) +1 (x-4) < 0`

`=> (x-4)(x+1) < 0`

`=> x in (-1,4)`

`=> x in (-1,4)`

But `x > 1`

`=> x in (1,4)`

Correct Answer is `=>` (C) `1 < x < 4`

Q 2418034800

If `log_(0.3)(x- 1) < log_(0.09)(x- 1)`, then `x` lies in
the interval
WBJEE 2016

(A)

`(2,oo)`

(B)

`(1,2)`

(C)

`(-2,-1)`

(D)

None of these

Solution:

Given, `log_(0.3) (x-1) < log_(0.09) (x-1)`

`=> log_(0.3) (x-1) < log_(0.3)^2 (x-1)`

`=> log_(0.3) (x-1)^2 < log_(0.3) (x-1)`

`=> (x-1)^2 > x -1` ` [ :. 0.3 < 1 ]`

`=> x^2 +1-2x -x +1 > 0`

`=> x^2 -3x +2 > 0`

`=> (x-1) (x-2) > 0`

`=> x < 1, x > 2 => x > 2` `[ :. x ` not less than `1]`

`:. (2, oo)`

Hence, `x` lies in the interval `(2, oo)`.

Correct Answer is `=>` (A) `(2,oo)`

Q 2365791665