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151.

Let `L`denote antilog_32 0.6 and M denote the number of positive integers whichhave the characteristic 4, when the base of log is 5, and N denote the valueof `49^((1-(log)_7 2))+5^(-(log)_5 4.)`Find the value of `(L M)/Ndot`

Answer» `L=antilog_32 0.6=(32)^(0.6)=(32)^(6/10)`
`L=8`
M from `5^4` to `5^5`
=625 to 3125
2500 integer
m=2500
N=`49^(1-log_7^2)+5^(-log_5^4)`
`=49*49^(-log_7^2)+5^(log_5^(4^(-1)`
`=49*7^(-2log_7^2)+4^(-1)`
`=49.2^(-2)+4^(-1)`
`=49*1/4+1/4`
`=49/4+1/4=50/4=25/2=N`
`(LM)/N=(8*2500*2)/25=1600`.
Discussion
152.

The number of positive integers satisfying `x+(log)_(10)(2^x+1)=x(log)_(10)5+(log)_(10)6`is...........

Answer» `x+log_10(2^x+1)=xlog_10 5+log_10 6`
`log_10 10^x=xlog_10 10=lambda`
`log_10 10^lambda+log_10(2^lambda+1)=log_10 5^x+log_10 6`
`2^x*5^x[2^x+1]=6*5^x=5^lambda[2^(2lambda(+2^lambda-6))]=0`
`5^x=0,2^(2lambda)+2^lambda-6=0`
`2^x=t^2+t-6=0`
`t^2+3t-2t-6=0`
`t(t+30-2(t+3)=0`
`(t-2)(t+3)=0`
`t=2,-3`
`2^lambda-2=0`
`lambda=1`.
Discussion
153.

If `(log)_(10)2=a ,(log)_(10)3=bt h e n(log)_(0. 72)(9. 6)` in terms of `a` and `b` is equal to (a) `(2a+3b-1)/(5a+b-2)` (b) `(5a+b-1)/(3a+2b-2)` (c) `(3a+b-2)/(2a+3b-1)` (d) `(2a+5b-2)/(3a+b-1)`

Answer» `log_0.72^9.6`
`log_10^9.6/log_10^0.72=(log_10^96-log_10^10)/(log_10^72-log_10^100)`
`=(log_10 (2^5*3)-1)/(log_10 (2^3*3^2)-2)`
`=(log_10 2^5+log_10 (3-1))/(log_10 2^3 +log_10 3^2 -2)`
`=(5a+b-1)/(3a+2b-2)`.
Discussion
154.

if `(a+log_4 3)/(a+log_2 3)= (a+log_8 3)/(a+log_4 3)=b` then find the value of `b`A. `1/2`B. `2/3`C. `1/3`D. `3/2`

Answer» Correct Answer - C
`(a+log_(4) 3)/(a+log_(2)3) = (a+log_(8) 3)/(a+log_(4)3) = (log_(4) 3-log_(8) 3)/(log_(2)3-log_(4) 3) = 1/3`
Discussion
155.

The value of `log ab- log|b|=`A. log aB. `log|a|`C. `-log a`D. none of these

Answer» Correct Answer - B
log ab is defined if ` ab gt 0` or a and b have the same sign.
Case (i): a, `b gt 0`
` rArr log ab - log|b| = log a + log b - log b = log a` …(i)
Case (ii): ` a, b lt 0`
` rArr log ab - log|b| = log (-a) + log (-b) - log (-b) `
` = log (-a)` ...(ii)
From Eqs. (i) and (ii), we have
` log ab - log|b| = log |a|`.
Discussion
156.

If `f(x)=log((1+x)/(1-x)),t h e n`(a)`f(x_1)f(x)=f(x_1+x_2)`(b)`f(x+2)-2f(x+1)+f(x)=0`(c)`f(x)+f(x+1)=f(x^2+x)`(d)`f(x_1)+f(x_2)=f((x_1+x_2)/(1+x_1x_2))`

Answer» `f(x_1)+f(x_2)=ln((1+x_1)/(1-x_1))+ln((1+x_2)/(1-x_2))`
`ln[((1+x_1)(1+x_2))/((1-x_1)(1-x_2))]`
`ln[(1+x_1+x_2+x_2x_1)/(1-x_1-x_2+x_1x_2)]-(1)`
`f((x_1+x_2)/(1+x_1x_2))=ln((1+(x_1+x_2)/(1+x_1x_2))/(1-((x_1+x_2)/(1+x_1x_2))))`
`=ln((1+x_1x_2+x_1+x_2)/(1+x_1x_2-x_1-x_2))-(2)`
`f(x)+f(x+1)=ln((1+x)/(1-x))+ln((1+x+1)/(1-x-1))`
`ln[((1+x)(2+x))/((1-x)x)]=ln((x^2+3x+2)/(x(1-x)))`
`f(x^2+x)=ln((1+x+x^2)/(1-(x^2+x)))=ln((x^2+x+1)/(1-x^2-x))`.
Discussion
157.

If a, b, c are consecutive positive integers and log (1+ac) = 2K, then the value of K isA. log bB. log aC. 2D. 1

Answer» Correct Answer - A
Let ` a = x - 1, b = x, c = x+1`
Now `log(1+ac)=log [1+(x-1)(x+1)]`
` = log x^(2) = 2 log x = 2 log b`
` rArr K = log b`
Discussion
158.

If `(a+(log)_4 3)/(a+(log)_2 3)=(a+(log)_8 3)/(a+(log)_4 3)=b` ,then b is equal to`1/2`(2) `2/3`(c) `1/3`(d) `3/2`

Answer» `(a+1/2log_2^3)/(a+log_2^3)=(a+1/3log_2^3)/(a+1/2log_2^3)`
`(a+t/2)/(a+t)=(a+t/3)/(a+t/2)`
`(a+t/2)^2=(a+t)(a+t/3)`
`a^2+t^2/4+at=a^2+(4at)/3+t^2/3`
`t/4+a=(4a)/3+t/3`
`-t/12=a/3`
`a=-t/4`
`b=(a+t/2)/(a+t)`
`=(-t/4+t/2)/(-t/4+t)`
`=(1/4)/(3/4)`
`=1/3`.
Discussion
159.

If `a , b , c`are consecutive positive integers and `(log(1+a c)=2K ,`then the value of `K`is`logb`(b) `loga`(c) 2(d) 1

Answer» a=m
b=m+1
c=m+2
`log(1+ac)=2k`
LHS
`=log(1+ac)`
`=log(1+m(m+2))`
`=log(1+m^2+2m)`
`=log(m+1)^2`
`=2log(m+1)`
`=2logb`
`2k=2logb`
`k=logb`.
Discussion
160.

Solve`(x-1)/(log_(3)(9-3^(x))- 3) le 1`.

Answer» We have `(x-1)/(log_(3)(9-3^(x))-3) le 1`
For this, we must have ` 9-3^(x) gt 0 or 3^(x) lt 9 or x lt 2`
The given expression can be expressed as:
` ((x-1))/(log_(3)(9-3^(x))-log_(3) 27) le 1 `
` rArr ((x-1))/(log_(3)((9-3^(x))/27))le 1`
` rArr (x-1)*log_(((9-3^(x))/27)) 3 le 1`
` rArr log_(((9-3^(x))/27))(3^(x-1)) lt 1`
As ` x lt 2, 0 lt (9-3^(x))/27 lt 1`
We have ` 3^(x-1) ge (9-3^(x))/27`
` rArr 9 xx 3^(x) ge 9 - 3^(x)`
` rArr 10 xx 3^(x) ge 9`
` rArr x ge log_(3) 0.9`
Therefore, ` x in [log_(3) 0.9, 2)`
Discussion
161.

Solve `log_(2)(2sqrt(17-2x))=1 - log_(1//2)(x-1)`.

Answer» Correct Answer - x = 4
`log_(2)(2sqrt17-2x)=1+log_(2)(x-1)`
` or log_(2)((2sqrt(17-2x))/(x-1))=1`
` or ((2sqrt(17-2x))/(x-1))=2`
`or 2sqrt(17-2x)=2(x-1)`
` or x^(2)2x+1 = 17 - 2x`
` x^(2) = 16`
` rArr x = 4 ("as "x ne -4)`
Discussion
162.

Solve` log_(2)(25^(x+3)-1) = 2 + log_(2)(5^(x+3) + 1)`.

Answer» Correct Answer - x = - 2
`log_(2)(25^(x+3)-1)=2+log_(2)(5^(x+3)+1)`
` or log_(2) (25^(x+3)-1)-log_(2)(5^(x+3)+1) = 2`
` or log_(2). (25^(x+3)-1)/(5^(x+3)+1) = 2`
`or (25^(x+3)-1)/(5^(x+3)+1) = 2^(2)`
` or y^(2) - 1 = 4y + 4" "("putting "5^(x+3)=y)`
` or y^(2) -4y-5 = 0`
` or y =-1, 5`
` rArr 5^(x+3) = 5 `
` or x =- 2`
Discussion
163.

`0.7625 xx 0.000357`

Answer» Correct Answer - 0.0002723
Discussion
164.

`69.13 xx 0.34 xx 0.014`

Answer» Correct Answer - 0.3291
Discussion
165.

Solve `log_(6) 9-log_(9) 27 + log_(8)x = log_(64) x - log_(6) 4`..

Answer» Correct Answer - ` x = 1/8`
`(log_(6)9+log _(6)4)-(log_(3)27)/(log_(3) 9) =(log_(8) x)/2 - log_(8) x`
` rArr 2 - 3/2 =- 1/2 log_(8) x`
` or 1/2 =- 1/2 log_(8) x`
` or x =1/8`
Discussion
166.

Solve ` 2 log_(3) x - 4 log_(x) 27 le 5`.

Answer» Let ` log_(3) x = y`
` rArr x = 3^(y)` (i)
Therefore, the given inequality becomes ` 2log_(x) x - 12 log_(x) 3 le 5`
` or 2y - 12/y le 5`
` or 2y^(2) - 5y - 12 le 0`
` or (2y+3)(y-4) le 0`
` rArr y in [-3/2 , 4]`
` rArr - 3/2 le log_(3) x le 4`
` rArr 3^(-3//2) le x le 81`
Discussion
167.

`358.6 xx 0.078 xx 0.5943`

Answer» Correct Answer - 16.63
Discussion
168.

Solve `log_(4)(2xx4^(x-2)-1)+4= 2x`.

Answer» Correct Answer - x = 2
` log_(4)(2xx4^(x-2)-1)+4 = 2x`
`or log_(4)(2xx4^(x-2)-1)=2x-4`
` or 2xx4^(x-2)-1 = 4^(2x-4)`
` or 2y-1 = y^(2)" "("putting "y=4^(x-2))`
` or y^(2)-2y+1 = 0`
` or y = 1`
` or 4^(x-2) = 1`
`or x = 2`
Discussion
169.

If ` a^(x) = b^(y) = c^(z) = d^(w)," show that " log_(a) (bcd) = x (1/y+1/z+1/w)`.

Answer» `log_(a)(bcd)=log_(a) b+log_(a) c+ log_(a) d`
Now` a^(x) = b^(y)`
` or (log b)/(log a) = x/y`
` or log_(a)b = x/y`
Similarly, ` log_(a) c= x/z and log_(a) d = x/w`
` rArr log_(a) (bcd) = log_(a) b+ log_(a) c+ log_(a) d`
`= x(1/y+1/z+1/w)`
Discussion
170.

Solve ` log_(4)(x-1)= log_(2) (x-3)`.

Answer» Correct Answer - x = 5
The given equality meaningful if
` x- 1 gt 0, x - 3 gt 0 rArr x gt 3`.
The given equality can be written as
` (log(x-1))/(log 4) =(log(x-3))/(log 2) `
` or log(x-1)= 2 log(x-3)(log 4 = 2 log 2)`
` or (x-1)=(x-3)^(2)`
` or x^(2) - 7x + 10 = 0`
` or (x-5)(x-2)=0`
` or x= 5 or 2`.
But ` x gt 3, so x = 5`.
Discussion
171.

Prove the following identities: (a) `(log_(a) n)/(log_(ab) n) = 1+ log_(a) b" "(b) log_(ab) x = (log_(a) x log_(b) x)/(log_(a) x + log_(b) x)`.

Answer» (a) `(log_(n) n)/(log_(ab) n) = (log_(n) ab)/(log_(n) a) = (log_(n) a+ log_(n) b)/(log_(n) a)`
` = 1+(log_(n) b)/(log_(n) a) = 1+log_(a) b`
(b) `(log_(a) xlog_(b) x)/(log_(a) x+log_(b) x) =(1/(log_(x) a)1/(log_(x) b))/(1/(log_(x) a)+1/(log_(x) b))=(1/(log_(x) a)1/(log_(x) b))/((log_(x) a + log_(x) b)/(log_(x) a log_(x) b))`
` = 1/(log_(x) a+log_(x) b)=1/(log_(x) ab)=log_(ab) x`
Discussion
172.

Solve:` 27^(log_(3)root(3)(x^(2)-3x+1) )=(log_(2)(x-1))/(|log_(2)(x-1)|)`.

Answer» Correct Answer - x = 3
We have `27^(log_(3)root(3)(x^(2)-3x+1) )=(log_(2)(x-1))/(|log_(2) (x-1)|)`
We must have `x-1 gt 0 and x - 1 ne 1`
` :. X gt and x ne 2`
If ` 1 lt x lt 2, log_(2) (x-1) lt 0`
`:. (log_(2)(x-1))/(|log_(2)(x-1)|) =- 1`
This is not possible as L.H.S. ` gt 0` always.
If ` x gt 2," then "(log_(2)(x-1))/(|log_(2)(x-1)|) = 1`
`:." Equation reduces to "27^(log_(3)root(3)(x^(2)-3x+1))=1`
`:gt x^(2)- 3x+1 = 1`
` rArr x = 0, 3," but "x ne 0`
` :. x = 3` is the only solution.
Discussion
173.

`(8.67 xx 99)/(1.78)`

Answer» Correct Answer - 19.19
Discussion
174.

Compute ` log_(ab)(root(3)a//sqrtb)" if " log_(ab) a = 4`.

Answer» Correct Answer - `17/16`
`log_(ab) a = 4`
` or 1/(log_(a)ab) = 4`
` or 1/(log_(a) a+log_(a) b) = 4`
` or 1+log_(a) b = 1/4`
` or log_(a) b=- 3/4`
Now` log_(ab) (root(3)a//sqrtb)=(log(root(3)a//sqrtb))/(log ab) = (1/3loga - 1/2 log b)/(log a+ log b) `
` (1/3-1/2 (logb)/(log a))/(1+(log b)/(log a))`
` = (1/3-1/2 log_(a) b)/(1+ log_(a) b)`
`= (1/3+1/2*3/4)/(1-3/4)`
` =(17/24)/(1/4)=17/6`
Discussion
175.

Find the value of`log_(1//3)root(4)(729*root(3)(9^(-1)*27^(-4//3)))`.

Answer» Correct Answer - -1
`log_(1//3) root(4)(729 xxroot(3)(9^(-1)xx27^(-4//3)))`
` = log_(1//3)root(4)(729 xxroot(3)(3^(-2)xx3^(-4)))`
` = log_(1//3) root(4)(3^(6)xx3^(-2))`
` = log_(1//3) 3`
` =-1`
Discussion
176.

`root(5)(8.0125)`

Answer» Correct Answer - 7.517
Discussion
177.

`(0.09634)^(3)`

Answer» Correct Answer - 0.0008941
Discussion
178.

The value of \(25^{\big(\frac{-1}{4}log_5\,25\big)}\) is(a) \(\frac{1}{5}\)(b) \(-\frac{1}{25}\)(c) –25 (d) None of these

Answer»

(a) \(\frac{1}{5}\)

\(25^{\big[\big(\frac{-1}{4}log_5\,25\big)\big]}\) = \(5^{\big[2\big(\frac{-1}{4}log_5\,25\big)\big]}\)

= \(5^{\big[\big(\frac{-1}{2}log_5\,25\big)\big]}\) = \(5^{log_5(25)^{-\frac{1}{2}}}\) = \(25^{-\frac{1}{2}}\) = \(\frac{1}{5}\)                      (∵ \(a^{log_ax}=x\))
Discussion