HackerRank Functions and Logarithm Quiz- 1

\log_{2}3125 =  \frac{\log3125}{\log2}

= \frac{\log5^{5}}{\log\frac{10}{5}}

=  \frac{5 \times \log5}{\log10 - \log5}

=  \frac{5 \times 0.698}{1 -  0.698}

=  \frac{3.49}{0.302}

= 11.56

It is a quadratic equation of the form ax^{2} + bx + c = 0  with a = 2, b = 5 and c = 3

b^{2} - 4ac  = 5^{2} - 4\times 2 \times 3

= 25 - 24 = 1 >0

b^{2} - 4ac  > 0

So f(x) have 2 distinct solutions and touches x axis at two points

\log{1.728} = 0.2375

\log{1.2^{3}} = 0.2375

3\times \log{1.2} = 0.2375

\log{1.2} =\frac{ 0.2375}{3}

= 0.0791

\frac{\log10}{\log\sqrt{10}} =  \frac{\log10}{\log(10^\frac{1}{2})}

= \frac{\log10}{\frac{1}{2} \times \log(10)}

= 2

f(x) exist only when x>0 and is -ve when 0<x<1

Let c is the reminder and (ax+b) is the quotient when  x^{2} + 5x - 2 is divided by (x + 2)

So, (x+2)(ax+b) + c =   x^{2} + 5x - 2

Simplifying,

ax^{2} + bx + 2ax + 2b + c =  x^{2} + 5x - 2

ax^{2} + (b + 2a)x + (2b + c) =  x^{2} + 5x - 2

Comparing,

a = 1,  b + 2a = 5 and 2b + c = -2

⇒ b = 3

And c = -8

f(x) = x^{4} - 4x^{3} + 2x^{2} + 7 = 0

f(-2) = (-2)^{4} - 4(-2)^{3} + 2(-2)^{2} + 7 = 0

= 16 + 32 +8 + 7 = 63

\log_{x}{1} = 0

⇒  x^{0} = 1

x can be any number

\log{a}^{b} = b \log{a}

So \log{4}^{27} = 27 \times \log{4} = 27 \times 2 \times \log{2}

= 27 x 2 x 0.301 = 16.254

Its Characteristic is 16

Number of digits = 16 + 1 = 17

\log_{3}{p} \times \log_{2}{9} = 2

\frac{\log{p}}{\log{3}} \times \frac{\log{9}}{\log{2}} = 2

\frac{\log{p}}{\log{3}} \times \frac{2 \times\log{3}}{\log{2}} = 2

\frac{\log{p}}{\log{2}} = 1

\log_{2}{p} = 1