IIT JEE , AIEEE Forum - Mathematics , Algebra

learner

log₃2, log₃ (2^x-5) , log₃(2^x -7/2) are in AP. Find x.

Kindly give solution or atleast hints.

Ans : 3

Rates assured !

Thanks! :-)

allamraju

They are in A.P $\Leftrightarrow$ (2^x-5)²=2(2^x-7/2) .put 2^x=k which gives k=4 or 8.But k=2^x>5 for log to be defined.so,k=8 and x=3

netkid07

m not writing base...coz..we don't need to alter it n all...so it's obviously 3....

log 2, log (2^x-5) , log (2^x -7/2) are in AP

2log (2^x-5)=log (2^x -7/2)+log 2

using properties of log...n taking antilog both sides, u have...

(2^x-5)^2 = 2(2^x - 7/2)

solve this quation for 2^x....then find the value of x...

hope it helps...cheero :)

learner

Thanks guys!! ..dunno why I keep overlooking simple stuff like that...

thanx a lot!!!

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They are in A.P(2x-5)2=2(2x-7/2) .put 2x=k which gives k=4 or 8.But k=2x>5 for log to be defined.so,k=8 and x=3

They are in A.P $\Leftrightarrow$ (2^x-5)²=2(2^x-7/2) .put 2^x=k which gives k=4 or 8.But k=2^x>5 for log to be defined.so,k=8 and x=3