Question

1. If 2log 4 + 2log 3 - 2log 12 = log x ,then x​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

3log2-2log3=1+log(1/x)

3log2-2log3=1+log(1/x)log 2^3 - log 3^2 = 1 + log (1/x)

3log2-2log3=1+log(1/x)log 2^3 - log 3^2 = 1 + log (1/x)log 2^3 - log 3^2 = log 10 + log (1/x)

3log2-2log3=1+log(1/x)log 2^3 - log 3^2 = 1 + log (1/x)log 2^3 - log 3^2 = log 10 + log (1/x)log 8 - log 9 = log 10 + log (1/x)

3log2-2log3=1+log(1/x)log 2^3 - log 3^2 = 1 + log (1/x)log 2^3 - log 3^2 = log 10 + log (1/x)log 8 - log 9 = log 10 + log (1/x)log (8/9) = log (10/x)

3log2-2log3=1+log(1/x)log 2^3 - log 3^2 = 1 + log (1/x)log 2^3 - log 3^2 = log 10 + log (1/x)log 8 - log 9 = log 10 + log (1/x)log (8/9) = log (10/x)8/9 = 10/x

3log2-2log3=1+log(1/x)log 2^3 - log 3^2 = 1 + log (1/x)log 2^3 - log 3^2 = log 10 + log (1/x)log 8 - log 9 = log 10 + log (1/x)log (8/9) = log (10/x)8/9 = 10/xx = 90/8

3log2-2log3=1+log(1/x)log 2^3 - log 3^2 = 1 + log (1/x)log 2^3 - log 3^2 = log 10 + log (1/x)log 8 - log 9 = log 10 + log (1/x)log (8/9) = log (10/x)8/9 = 10/xx = 90/8x = 11.25

Step-by-step explanation:

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