find x in the following question..
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9^2(5/x) = 343
9^10/x = 7^3
taking log on both the sides
(10/x)log3^2 = 3log7
20/xlog3 = 3 log 7
(20/x ) log 3= 3 × log 7
9.54/2.53 = x
= 3.77
9^10/x = 7^3
taking log on both the sides
(10/x)log3^2 = 3log7
20/xlog3 = 3 log 7
(20/x ) log 3= 3 × log 7
9.54/2.53 = x
= 3.77
eshitajain:
5/x is exponent
then u have to mention it by using ^
ok
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