Question

Prove : Log 2 base 2 + log 2 base 8- log 2 base 4 -log 2 base 16 = 7/12

Answer 1

log 2 base 2 = 1

log 2 base 8 = 1/3

log 2 base 4 = 1/2

log 2 base 16 = 1/4

1+1/3-1/2-1/4 = 7/12

Answer 2

Topic :-

Logarithms

Given to prove that :-

$log2_2 + log2_8 - log2_4 - log2_{16} = \dfrac{7}{12}$

Formulae used :-

$loga_a = 1$

$log a_{a^n} = \dfrac{1}{n } loga_a\\=\dfrac{1}{n}$

Proof :-

Take L.H.S

8 can be written as 2×2×2

8 = 2³

4 can be written as 2×2

4 = 2²

16 can be written as 2×2×2×2

16 = 2⁴

$log2_2 + log2_8 - log2_4 - log2_{16}$

$(1) + log2_{2^3} - log2_{2^2} - log2_{2^4}$

$1 + \dfrac{1}{3} (log2_2) -\dfrac{1}{2} (log2_2) - \dfrac{1}{4} (log2_2)$

$1+\dfrac{1}{3} (1)-\dfrac{1}{2} (1)-\dfrac{1}{4} (1)$

$1+\dfrac{1}{3} -\dfrac{1}{2} -\dfrac{1}{4}$

Take L.C. M to the denominator

L.C.M of 2,3,4 is 12

$\dfrac{12+1(4)-1(6)-1(3)}{12}$

$\dfrac{12+4-6-3}{12}$

$\dfrac{16-9}{12}$

$\dfrac{7}{12}$

So,

$\red{\boxed{log2_2 + log2_8 - log2_4 - log2_{16} = \dfrac{7}{12}}}$

Hence proved !

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