Logarithms Quiz-1

\log_{2}512

=\log_{2}2^{9}

=9\log_{2}2                                                         

=9 x 1

=9

\log_{3}x = 4     

    x = 3^4 .

\log_{x}y=22

y=x^22

y=(3^{4})^{22}     [put value of x]

y=3^{88}

\left ( \frac{\log a}{\log \sqrt{a}} \right )^3

=\left ( \frac{2\log a}{\log a} \right )^3

=2^3

=8

 \log_{10}(\frac{1}{80})

=\log_{10}1 - \log_{10}80

= -\log_{10}(8\times 10)

=-(\log_{10}8 + \log_{10}10)

  = - (p + 1)

\log_{9}10

= \frac{1}{\log_{10} 9} 

= 1/0.9542

= 10000/9542

= 1.047

\frac{1}{\log_{5} 60}+\frac{1}{\log_{3} 60}+\frac{1}{\log_{4} 60}

=\log_{60} 5+\log_{60} 3+\log_{60} 4

=\log_{60} (5\times 3\times 4)

=\log_{60} 60

=1

\log_{2} (pq)^2

=2\log_{2} (pq)

=2(\log_{2} p+\log_{2} q)

\log_{2} (p+q)=\log_{2} 10

=> p+q = 10.......................(i)

\log_{2} (pq)=4

=>pq = 2^4 =16..................(ii)

So, from (i) and (ii) we can get the value of p-q using below formula

(p-q)2 = (p+q)2 - 4pq

=>(p-q)2 = (10 x 10) - (4 x 16)

=>(p-q)2 =100 - 64

=> (p-q)2 = 36

=>p-q = 6

\log {7}^{20} = 20 * log 7

=20 * 0.845

= 16.90

Therefore, the number of digits in 7²⁰ = 17 (approx.)

log (1/p) + log p + log (1/q) + log q = log (p+q)

=>log (p x q)/(p x q) = log (p+q)

=>log 1 = log (p+q)

=> p+q = 1