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thx alot guys

I think that, on the second question, he means "log base 3 of 3 + log base 9 of 1".If so, it's easy. 3¹ = 3, and 9^0 = 1. So, 1 + 0 = 1.

1) xlog4 + log7 = xlog9+ log5xlog4 - xlog9 = log5 - log7xlog9 - xlog4 = log7 - log5x(log9 - log4) = log7 - log5x = (log7 - log5)/(log9 - log4)The second question isn't clear

hi can someone explain in detail how to solve this kind of problem please:

1) 7*4^x = 5*9^x 2)log3 3 + log9 1

you should use logarithms to solve it, thx!