Logarithm

Algebra Level 1

Let j and p be the roots of $x^{2}-3x +1=0$ . Find the value of $log_{2}$ j + $log_{2}$ p .

2 solutions

Saurabh Mallik
May 12, 2014

Firstly we need to find the roots of $x^2-3x+1=0$

By using quadratic equation:

$= \frac{-(-3)+-\sqrt{(-3)^{2}-4\times1\times1}}{2\times1}$

$= \frac{3+-\sqrt{9-4}}{2}$

$= \frac{3+-\sqrt{5}}{2}$

So, $j=\frac{3+\sqrt{5}}{2}$ and $p=\frac{3-\sqrt{5}}{2}$

So, solving: $log_{2}j+log_{2}p$

$= log_{2}(\frac{3+\sqrt{5}}{2})+log_{2}(\frac{3-\sqrt{5}}{2})$

$= log_{2}(\frac{3+\sqrt{5}}{2})(\frac{3-\sqrt{5}}{2})$

$= log_{2}(\frac{(3+\sqrt{5})(3-\sqrt{5})}{2\times2})$

$= log_{2}(\frac{3^{2}-(\sqrt{5})^{2}}{4})$

$= log_{2}(\frac{9-5}{4})$

$= log_{2}(\frac{4}{4})$

$= log_{2}1=0$ since $2^{0}=1$

So, the answer is: $log_{2}1=\boxed{0}$

log2j +log 2p= log2 (j*p)=(product of roots of quadratic equation)c/a=1/1=1 log2 1=0

manas vema - 7 years ago

2 is the base

manas vema - 7 years ago

Manas please vote my solutions. My solutions are much better than others.

Saurabh Mallik - 7 years ago

To include base in your solutions, refer to the 'Formatting guide'.

To include base, type: log_{a} with brackets.

Example: log_{2} = $log_{2}$

Saurabh Mallik - 7 years ago

So lengthy

Chinmoyranjan Giri - 6 years, 11 months ago

Divyam Bapna
May 1, 2014

i p=1 log(i p)=log1=0