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

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

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

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