Solving logarithms algebraically

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log(base17)17^x+2=log(base7)1

Apr 29th, 2015

we know log(base a)(b)= ln(b)/ln(a)  and also  a^x =  e^(xln(a))

So :  log(base 17)(17^x+2) = ln(17^x + 2)/ln(17) 

and  log(base 7) 1 = ln(1)/ln(7) =  0 because  ln(1) = 0

so ln(17^x + 2)/ln(17) =  0  =>  ln(17^x + 2) = 0   =>  ln(17^x + 2) = ln(1)  => 17^x +2 = 1  

So :   e^(xln(17)) = -1   =>    e^(xln(17)) = -1 < 0  and we know thats  e^(any number) > 0  always 

So we don't have any solutions  

  Assuming  we have  log(base17)17^x - 2=log(base7)1  the solution is : e^(xln(17)) = 3 = e^(ln(3)) => xln(17)=ln(3) so x=ln(3)/ln(17) 


SO GOOD LUCK  and please if you don't understand something please write it in the comment 

please if you have any other homework please call me first i want to be the best tutor 

Apr 29th, 2015

Yeah, the part that get's me is solving equations algebraically. 

Apr 30th, 2015

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