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Question: If \(x^{2}\) then \(A + B = 1 \Rightarrow A = 1 - B = 1 - 0 = 1\)...

If x2x^{2} then A+B=1⇒A=1−B=1−0=1A + B = 1 \Rightarrow A = 1 - B = 1 - 0 = 1

A

2x=A(x2+x+1)+(Bx+C)(x−1)x=12x = A(x^{2} + x + 1) + (Bx + C)(x - 1)x = 1

B

2=3A2 = 3A

C

⇒\Rightarrow

D

None of these

Answer

2=3A2 = 3A

Explanation

Solution

log⁡7log⁡7777=log⁡7log⁡777/8=log⁡7(7/8)\log_{7}{\log_{7}\sqrt{7\sqrt{7\sqrt{7}}}} = \log_{7}{\log_{7}7^{7/8}} = \log_{7}(7/8) =log⁡77−log⁡78=1−log⁡723=1−3log⁡72= \log_{7}7 - \log_{7}8 = 1 - \log_{7}2^{3} = 1 - 3\log_{7}2 81(1/log⁡53)+27log⁡936+34/log⁡7981^{(1/\log_{5}3)} + 27^{\log_{9}36} + 3^{4/\log_{7}9}

=34log⁡35+33.12log⁡336+34log⁡97= 3^{4\log_{3}5} + 3^{3.\frac{1}{2}\log_{3}36} + 3^{4\log_{9}7} =3log⁡354+3log⁡3363/2+3log⁡374/2= 3^{\log_{3}5^{4}} + 3^{\log_{3}36^{3/2}} + 3^{\log_{3}7^{4/2}}

=54+363/2+72=890=log⁡(167157.255245.813803)=log⁡2ab=log⁡45.log⁡56=log⁡46=12log⁡26= 5^{4} + 36^{3/2} + 7^{2} = 890 = \log\left( \frac{16^{7}}{15^{7}}.\frac{25^{5}}{24^{5}}.\frac{81^{3}}{80^{3}} \right) = \log 2ab = \log_{4}5.\log_{5}6 = \log_{4}6 = \frac{1}{2}\log_{2}6.