Logarithms Practice Test 2

Question : 10 of 39

Marks: +1, -0

\log 3, \log (3^x - 2)

and

\log (3^x + 4)

are in arithmetic progression, then x is equal to:

Solution:

In arithmetic progression, common ratios are equal. So

\log(3^x-2)-\log3=\log(3^x+4)-\log(3^x-2)

⇒{\log(3^x-2)}/{\log3}={\log(3^x+4)}/{\log(3^x-2)}

⇒{\log 3^x }/{\log2\log3}={x\log 3 \log4 \log2}/{x\log 3 }

⇒{x\log 3 }/{\log2\log3}={x\log 3 \log4 \log2}/{x\log 3 }

⇒x/{\log 2}=\log 4\log2

x = \log 4 \log 2 \log 2

x=\log 8

x = \log 2^3

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