Question

If 2log (base 8) N=P, log (base 2) 2N=Q and Q-P=4, find N

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Answer 1

Step-by-step explanation:

P = 2 log N base 8

= 2× log N/ log 8

= 2× log N/ log 2³ = 2× log N / 3 log 2

= 2/3 * log N base 2

log 2N base 2 = Q

let log x base 2 be represented as log x, where x is any no. or variable

Q - P = 4

log 2N - 2/3* log N = 4

log 2N - log N ^⅔ = 4

log 2N/N^⅔ =4

log 2× N ^(1-⅔) = 4

log 2× N^⅓ =4

2×N ^⅓ = 2⁴ as log base = 2

N^⅓ = 2⁴/2

N ^ ⅓ = 2³

N = 2^(3×3) = 2⁹

Answer 2

Given :

$\sf \bullet \: \: 2 log_{8}(N) = \: P \\ \\ \sf \bullet \: log_{2}(2 N ) \: = Q \\ \\ \sf \bullet \: Q \: - \: P = 4$

To find : N

Identity used :

$\sf \bullet \: log_{a}(b) = \dfrac{ log_{x}(a) }{ log_{x}(b) } \\ \\ \sf \bullet \: log(ab) = log(a) \: + \: log(b) \\ \\ \sf \bullet \: log( {a}^{m} ) \: = \: m \: \times log(a) \\ \\ \:\sf \bullet log_{a}(a) = 1 \\ \\ \sf \: \bullet \: if \: log_{x}(a) = log_{x}(b) \\ \sf \: then \: \: \: a = b$

Solution :

$\sf \: Q - P = 4 \\ \\ \sf \implies \: log_{2}(2 N) \: - \: 2 \{ \: log_{8}(N) \} \: = 4 \\ \\ \: \sf \implies \: log_{2}(2N) \: - \: 2 \bigg \{ \: \dfrac{ log_{2}(N) }{ log_{2}(8)} \bigg \} \: = 4 \\ \\ \sf \implies \: log_{2}(2N) \: - \: 2 \bigg \{ \: \dfrac{ log_{2}(N) }{ log_{2}( {2}^{3} )} \bigg \} \: = 4 \\ \\ \sf \implies \: \bigg \{ log_{2}(2) + log_{2}( N ) \bigg \} \: - \: 2 \bigg \{ \: \dfrac{ log_{2}(N) }{3 \times log_{2}(2)} \bigg \} \: = 4 \\ \\ \sf \implies \: \bigg \{ 1 \: + log_{2}( N ) \bigg \} \: - \: 2 \bigg \{ \: \dfrac{ log_{2}(N) }{3 \times 1} \bigg \} \: = 4 \\ \\ \sf \implies \: 1 \: + log_{2}( N ) - \: \: \dfrac{ 2 }{3 }log_{2}(N) \: = 4 \\ \\ \sf \implies \: log_{2}( N ) \bigg \{ \: 1 - \dfrac{2}{3} \bigg \} \: = 4 - 1 \\ \\ \sf \implies \: log_{2}( N ) \bigg \{ \dfrac{1}{3} \bigg \} \: = 3 \\ \\\sf \implies \: log_{2}( N ) = 3 \times 3 \\ \\ \sf \implies \: log_{2}( N ) =9 \times log_{2}(2) \\ \\ \sf \implies \: log_{2}( N ) = log_{2}( {2}^{9} ) \\ \\ \sf \implies \: N \: = \: {2}^{9} = 512$

ANSWER :

N = 2⁹ = 512