Logarithm and its uses - class-XI

$\log _ee^5=5\log _ee=5*1=5$

We have,

${{\log } _{3}}x=3$

We know that

${{\log } _{a}}x=y\Rightarrow x={{a}^{y}}$

Therefore,

$ x={{3}^{3}} $

$ x=27 $

Here,

$ {{\log } _{x}}y=12 $

$ y={{x}^{12}} $

On putting the value of $x$, we get

$y={{3}^{12}}$

Hence, this is the answer.

$antilog _{10}(0.3678)=2.3324$

$\log _264+\log _3729\\log _22^6+\log _33^6\6\log _22+6\log _33\6+6=12$

$\log _5625+\lg _6216\\log _55^4+\log _66^3\4\log _55+3\log _66\3+4=7$

We have, 

$\log x=-2.0258$

The number before decimal point is $2,$ so decimal point will be after $3$ digits.

The number before decimal point is $0,$ so decimal point will be after $1$ digits.

The value of $[\log _{10}(5\log _{10} 100)]^{2}$ is
$=[\log _{10}(5\log _{10}10^2)]^2$

From Logarithmic table,

From logarithmic table,

From log table,

From Logarithmic table,

$\log(x+1)=2\log x$

Here, $g^{\log _{3}(\log _{2}x)}$ is an exponential function and $\log _{2}{x}-\left(\log _{2}{x}\right)^{2}+1$ is a quadratic with imaginary roots.

$2y=\log\left(12-5x-3x^2\right)$

Let $x= \dfrac{(17.42)^{2/{3}}\times 18.42}{\sqrt{126.37}}$
Taking logarithm on both sides,
$ \log { x } =\log { (17.42)^{ 2/{ 3 } } } +\log { 18.42 } -\log { \sqrt { 126.37 }  } $

$\log { x } =\dfrac { 2 }{ 3 } \log { 17.42 } +\log { 18.42 } -\dfrac { 1 }{ 2 } \log { (126.37) } $

$ \log { x } =\dfrac { 2 }{ 3 } \log { (1.742\times 10) } +\log { (1.842\times 10) } -\dfrac { 1 }{ 2 } \log { (1.264\times { 10 }^{ 2 }) } $

$ \log { x } =\dfrac { 2 }{ 3 } { (1.2410) } + { 1.2653 } -\dfrac { 1 }{ 2 } { (2.1018) } $

$\log { x } =0.8273+1.2653-1.0509$

$\log { x } =1.0417$
$\Rightarrow x= \text{antilog }(1.0417)$
$\Rightarrow x = 11.01$

Given a = log3(log3(2)). The inequality 1 < 2^(-k + 3^-a) < 2 simplifies to 0 < -k + 3^-a < 1. Since 3^-a = 3^-log3(log3(2)) = 1/log3(2) = log2(3) approx 1.585. So 0 < -k + 1.585 < 1, which means 0.585 < k < 1.585. Thus k must be less than 1.585.

$a=\log _{ 3 }{ 5 } =\cfrac { \log { 5 }  }{ \log { 3 }  } ,b=\cfrac { \log { 25 }  }{ \log { 7 }  } =\cfrac { \log { 5^2 }  }{ \log { 7 }  }=\cfrac { 2\log { 5 }  }{ \log { 7 }  } $

The equation simplifies significantly because log_x(x) = 1. The equation becomes x^(3/4 * 1^2) + 1 - 5/4 = sqrt(2), which is x^(3/4) - 1/4 = sqrt(2). This has a real solution for x.

$\log _{10}1016\Rightarrow \dfrac{\log 1016}{\log 10}$

$\log _{7}\log _{5}(\sqrt{x^{2}}+5+x)=0$

log(1.5) = log(3/2) = log(3) - log(2) = b - a.

The derivative y' = a/x + 2bx + 1. Setting y' = 0 at x = 2 and x = -4/3 gives a system of equations: a/2 + 4b + 1 = 0 and a/(-4/3) - 8b/3 + 1 = 0. Solving these yields a=12 and b=-10.

Using log properties and the identity (x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3, the expression simplifies to a value that results in 2^A = 72.

We have to find $\log (0.002359)$

log4(256) = 4. Then log2(log2(4)) = log2(2) = 1. So A = 1 + 2 * log_sqrt(2)(2) = 1 + 2 * 2 = 5.

Let,  $y=\displaystyle \log _{\frac{1}{20}}40$
$\Rightarrow y=-\displaystyle \log _{20}40 \quad [\because \log _{1/a}b=-\log _ab]$
Now $20 < 40<20^2\Rightarrow \log _{20}20<\log _{20}40<\log _{20}20^2$
$\Rightarrow 1< \log _{20}40<2 \quad [\because \log a^m=m\log a, \log _aa=1]$
$\Rightarrow 1<-y<2\Rightarrow -2<y<-1$

$\displaystyle \frac { 2 }{ 3 } <\frac { 5 }{ 6 } <1$
Taking log with base $\displaystyle \frac { 2 }{ 3 } $
$\displaystyle \log _{ \frac { 2 }{ 3 }  }{ \frac { 2 }{ 3 }  } >\log _{ \frac { 2 }{ 3 }  }{ \frac { 5 }{ 6 }  } >\log _{ \frac { 2 }{ 3 }  }{ 1 } $  (Since the base $\displaystyle \frac { 2 }{ 3 } <1)$
$\displaystyle \Rightarrow 1>\log _{ \frac { 2 }{ 3 }  }{ \frac { 5 }{ 6 }  } >0$

$log _a 1 = m$
$\Rightarrow a^m = 1 a^0    \Rightarrow m = 0$

Now, $1+x=\log _{ a }{ a } +\log _{ a }{ bc } =\log _{ a }{ abc } $
$\Rightarrow \dfrac { 1 }{ 1+x } =\log _{ abc }{ a } $
Similarly, $\dfrac { 1 }{ 1+y } =\log _{ abc }{ b } $ and $\dfrac { 1 }{ 1+z } =\log _{ abc }{ c } $
$\therefore \dfrac { 1 }{ 1+x } +\dfrac { 1 }{ 1+y } +\dfrac { 1 }{ 1+z } $
                    $=\log _{ abc }{ a } +\log _{ abc }{ b } +\log _{ abc }{ c } $
                    $=\log _{ abc }{ abc } =1$

If a number is $N>0$, then $\log _{10}N$ will have two parts, the integral part is known the characteristic and the decimal part is known as mantissa.

Given, $x= \log _{ 12 }{ 640 } $

Given, $\displaystyle { log } _{ 5 }{ log } _{ 5 }{ log } _{ 2 }x=0$
$\displaystyle \Rightarrow \quad { log } _{ 5 }{ log } _{ 2 }x=50$
$\displaystyle \Rightarrow \quad { log } _{ 5 }{ log } _{ 2 }x=1\Rightarrow { log } _{ 2 }x=5$
$\displaystyle \Rightarrow x={ 2 }^{ 5 }=32$

$\log _{10}0.0006024$

$\log _{10}0.5432$

Antilog $\bar1.8840$

The number before decimal point is $2,$ so decimal point will be after $3$ digits.

The number before decimal point is $1,$ so decimal point will be after $2$ digits.

The number before decimal point is $0,$ so decimal point will be after $1$ digits.

For number less than $1,$ if there are $n$ zeroes after the decimal point, then the characteristics of that number will be $-(n+1).$

$\log { y } =\cfrac { 1 }{ 3 } \left[ \log { (16.23) } -\log { (426.8) }  \right] \ \log { y } =\cfrac { 1 }{ 3 } \left[ 1.2103-2.63022 \right] \ \log { y } =-0.4733\ y={ 10 }^{ -0.4733 }=0.3362$

Let,$f(x)=y=\log x$

The number before decimal point is $4,$ so decimal point will be after $5$ digits.

$\log _{10}8=\log _{10}2^3=3\log _{10}2$

We know that $AntiLog(x)=10^{x}$

Given $x=(0.15)^{20}$

Given that:

$\log _{2}{\cfrac{625}{10000}}$

$y=(0.036)^{16}$

$\log _{10}{2}=a$

The expression is 4log10(x) - log2(10^-4) = 4log10(x) + 4*log2(10). This does not have a simple maximum for 0 < x < 1 as it approaches -infinity. The question is likely garbled.

Let there be $N$ integers whose characteristic is 3, when base of log is 7
Then, $\log _{ 7 }{ N } =x$ where $3\le x<4$
As $3\le x<4$
$3\le \log _{ 7 }{ N } <4\ \Rightarrow { 7 }^{ 3 }\le N<{ 7 }^{ 4 }\ \Rightarrow N={ 7 }^{ 4 }-{ 7 }^{ 3 }\ \Rightarrow N=2401-343=2058$ 

Let $y=\dfrac { 28.45\times \sqrt [ 3 ]{ .3254 }  }{ 32.43\times \sqrt [ 3 ]{ .3046 }  } $

Given $\log _{ 10 }{ 2 } =0.301$

The number of zeroes will be
$=|log _{10}(3^{-100})|$
$=|-100(log _{10}(3))|$
$=|-47.71|$
$=47.71$
Taking the integral part (since number of zeroes has to be an integer, there will be $47$ zeros.

Approximate value of $\log _{ 11 }{ 21 } $