How do you solve this log problem?

GamaPro1

From: US
Posts: 19
Votes: 38
Solve for x: Log(x^35)=log(2x)+log(9^2) I noticed the bases aren't the same and that nothing can be factored out. I also thought of making it into a quadratic but that wouldnt work as x would be to the 35th power. Thanks

 

Answer No.1

WizTeacher

From: US
Posts: 18
Votes: 54
X^35 = 2x(9^2) Cancel x, and then solve for x, x = 162^(1/34) = 1.1614

Answer No.2

Samuel89jack

From: US
Posts: 37
Votes: 74
Umm...are we looking at the same problem? The way it's written here, there is no difference between the bases (all are log base 10). Also, remember that log(a^b) = bloga and log(b)+log(a)=log(ab). 35logx=log(2x)+log(81) 35logx=log(162x) 10^(35logx) = 10^(log(162x)) Remember: 10^logx=x and x^(ab)=(x^a)^b (10^logx)^35 = 10^log(162x) x^35=162x x^34=162 x=34rt(162) (the 34th root of 162) x~~1.022

Answer No.3

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