Tag: introduction to logarithm

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 } $

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

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