Logarithmic notation - class-XI

We have,

$ \log \left( -2x \right)=2\log \left( x+1 \right) $

$ \Rightarrow \log \left( -2x \right)=\log {{\left( x+1 \right)}^{2}} $

Comparing both side and we get,

$ -2x={{\left( x+1 \right)}^{2}} $

$ \Rightarrow -2x={{x}^{2}}+1+2x $

$ \Rightarrow {{x}^{2}}+4x+1=0 $

Using quadratic formula and we get,

$ x=\dfrac{-4\pm \sqrt{16-4\times 1\times 1}}{2\times 1} $

$ x=\dfrac{-4\pm \sqrt{12}}{2} $

$ x=\dfrac{-4\pm \sqrt{2\times 2\times 3}}{2} $

$ x=\dfrac{-4\pm 2\sqrt{3}}{2} $

$ x=-2\pm \sqrt{3} $

Hence,
this is the answer.

$x^y=z$
Apply log on both the sides, we get
$\therefore ylogx=logz$
$\therefore y=\dfrac{logz}{logx}$
$\therefore y=log _xz$
Hence, it is false.

$\log _{ 10 }{ 0.01= } \log _{ 10 }{ { 10 }^{ -2 } } $

$ \log _{ 10 }{ 1 }= 0\ 1 = { 10 }^{ 0 }$

$ \log _{ 5 }{ 125 } =x\ 125 = { 5 }^{ x }\ x = 3 $

$ \log _{ 5 }{ 1 } =x\ 1\quad =\quad { 5 }^{ x }\ x\quad =\quad 0\ \ $

$ \log _{ 10 }{ 0.01 } = -2 \Rightarrow 0.01 = { 10 }^{ -2 } $

$y={ a }^{ x }\Rightarrow \log _{ a }{ y } =x\ \therefore 81={ 3 }^{ 4 }\Rightarrow \log _{ 3 }{ 81 } =4$

$log 27 = 1.431$
$\Rightarrow log (3^3) = 1.431$
$\Rightarrow 3 log 3 = 1.431$
$\Rightarrow log 3 = 0.477$
$\therefore log 9 = log (3^2) = 2 log 3 = (2 \times 0.477) = 0.954$

Given, $\log _c a=x$

(a) Since $log _a a = 1,$ so $log _{10}  10 = 1$
(b) $log (2 + 3) log 5$ and $log (2 \times 3) = log 6 = log 2 + log 3$
$\therefore log(2 + 3) \neq log (2 \times 3)$
(c) Since, $log _a  1 = 0$, so, $log _{10} 1 = 0$.
(d) $log(1 + 2 + 3) = log 6 = log (1 \times 2 \times 3) = log 1 + log 2 + log 3$

$log _{10} 80 = log _{10} (8 \times 10)$
$= log _{10} 8 + log _{10} 10$
$=log _{10} (2^3) + 1$
$= 3 log _{10} 2 + 1$
$= (3 \times 0.3010) + 1$
$= 1.9030$

$log _5 512 = \dfrac{log 512}{log 5}$
$=\dfrac{log 2^9}{log (10/2)}$
$=\dfrac{9 log  2}{log 10 - log 2}$
$=\dfrac{9 \times 0.3010}{1- 0.3010}$
$= \dfrac{2.709}{0.699}$
$=\dfrac{2709}{699}$
$= 3.876$

$log (2^{64}) = 64 \times log 2$
$= (64 \times 0.30103)$
$= 19.2592$
Its characteristic is 19.
Hence, then number of digits in $2^{64}$ is 20.

$log _x \left( \dfrac{9}{16} \right ) = - \dfrac{1}{2}$
$\Rightarrow x^{-1/2} = \dfrac{9}{16}$
$\Rightarrow \dfrac{1}{\sqrt x} = \dfrac{9}{16}$
$\Rightarrow \sqrt x = \dfrac{16}{9}$
$\Rightarrow x = \left( \dfrac{16}{9} \right)^2$
$\Rightarrow x = \dfrac{256}{81}$

The value of $\dfrac {1}{2}\log _{10} 25 - 2 \log _{10} 3 +\log _{10} 18$ is
$= \log _{10}(25)^{1/2} - \log _{10} (3)^{2} + \log _{10}18$
$= \log _{10}5 - \log _{10}9 + \log _{10}18$
$= \log _{10} \left (\dfrac {5}{9}\times 18\right ) $

Let $log _2      16 = n$
Then, $2^n = 16 = 2^4    \Rightarrow n = 4$
$\therefore log _2  16 = 4$

$5^2=25$

Exponential form of $\log _2 16=4$

Mantissa of logarithm of $719.3$ to base $10$ is $0.8569$

The 6th term of the binomial expansion (a+b)^7 is 7C5 * a^2 * b^5. Solving the equation for x leads to x=2.

We have,

$ x={{\log } _{3}}243\,\,......\,\,\left( 1 \right) $

$ y={{\log } _{2}}64\,\,......\,\,\left( 2 \right) $

Applying rule,

${{\log } _{b}}x=y\,\Rightarrow {{b}^{y}}=x$

So,

From equation (1) and (2) to,

$ x={{\log } _{3}}243 $

$ \Rightarrow {{3}^{x}}=243 $

$ \Rightarrow {{3}^{x}}={{3}^{5}} $

$ \Rightarrow x=5 $

Now,

$ {{\log } _{2}}64=y $

$ \Rightarrow {{2}^{y}}=64 $

$ \Rightarrow {{2}^{y}}={{2}^{6}} $

$ \Rightarrow y=6 $

Now,

$ \sqrt{x-2\sqrt{y}} $

$ =\sqrt{5-2\sqrt{6}} $

Hence, this is the answer.

The exponential form 3*sqrt(8) = 2 is not standard. However, if the expression is 8^(1/3) = 2, then the logarithmic form is log_8(2) = 1/3. The provided option A matches this logic.

Solving the logarithmic equation requires using properties of logs and substitution. The resulting value of x satisfies the inequality x > 10.

$5^3=125$
Taking log on both sides, we get
$3\log 5 = \log 125$
$\log _5 125 = 3$         ...(since $\dfrac{\log a}{\log b} = \log _ba$)

$81^{\frac{3}{4}}=27$
Taking log on both sides
$\dfrac{3}{4}log81=log27$
$log _{81}27= \dfrac{3}{4}$.....(since $\dfrac{loga}{logb}=log _ba$)

$3^{-2}=\dfrac{1}{9}$
Taking log on both sides
$-2log3= log\dfrac{1}{9}$
$log _3\dfrac{1}{9}= -2$.....(since $\dfrac{loga}{logb}=log _ba$)

$10^{-3}=0.001$
Taking log on both sides
$-3log10= log0.001$
$log _{10}0.001= -3$.....(since $\dfrac{loga}{logb}=log _ba$)

$\displaystyle \log : V = 2 : \log : 2 - : \log : 3 + : \log : \pi + 3 : \log : r$
$\therefore \log V = \log 2^2-\log 3+\log \pi+ \log r$....($\log a^b=b \log a$)
$\therefore \log V= \log \cfrac{4}{3}\pi r^3$.....($\log a.b = \log a + \log b, \log \cfrac {a}{b} = \log a - \log b$)
$\therefore V=\cfrac{4}{3}\pi r^3$

$ \ \log _{ 10 }{ 0.001 } =x\ 0.001 = { 10 }^{ x }\ \dfrac { 1 }{ 1000 } = { 10 }^{ -3 }\ x=-3  $

$ \log _{ 0.5 }{ 16 } = x $

$0.001$ to the base $10$
So, as per logarithm,
$\log _{ 10 }{ 0.001 } =\log _{ 10 }{ { 10 }^{ -3 } } $
$\log _{ 10 }{ 0.001 } =-3\log _{ 10 }{ { 10 } } $
$\log _{ 10 }{ 0.001 } =-3\times 1$
$\log _{ 10 }{ 0.001 } =-3$

Let us assume that $\log _{ 2 }{ 5 } =\frac { p }{ q } $ , where $p,q$ are integers

$ \log _{ 10 }{ (x-10) } =1$

We have,

Consider the given equation.

We have,

Let $\log _{\frac{1}{2}}4 = x $

Then, $(\dfrac{1}{2})^x = 4 $

$Or, (\dfrac{1}{2})^x =2^2$

$Or, 2^{-x} = 2^2$

$x= -2$

Let $ \log _{ 10 }{ x+2 } =a$ and $ \log _{ 10 }{ x-1 } =b$
$\therefore a+b=2\log _{ 10 }{ x+1 } $ (from the question)
Thus, the given equation(in the question) reduces to ${a}^{3}+{b}^{3}={(a+b)}^{3}$
$\Rightarrow 3ab(a+b)=0$
$\Rightarrow a=0$ or $b=0$ or $a+b=0$
$\Rightarrow \log _{ 10 }{ x+2 }=0$  or $\log _{ 10 }{ x-1 }=0$ or $2\log _{ 10 }{ x } +1=0$
$\Rightarrow x={10}^{-2}$  or  $x=10$ or  $x={ 10 }^{ -\frac { 1 }{ 2 }  }$
Hence  $x=\left{ \dfrac { 1 }{ 100 },10 ,\dfrac { 1 }{ \sqrt { 10 }  }  \right} $