Question

Solve the following equation analytically. 2 log9(x) = log9(2) + log9(x + 40) (the 9's are base 9)

Answer 1

\[2\log_9(x)=\log_9(2)+\log_9(x+40)\]

Answer 2

\[2\log_{9}(x) = \log_{9}(2) + \log_{9}(x+40)\]
\[\log_{9}(x^{2}) = \log_{9}(2(x+40))\]
\[\log_{9}(x^{2}) = \log_{9}(2x+80)\]
x^2-2x-80
(x-10)(x+8)
x=10 or -8

Answer 3

\[\log_9(x^2)=\log_9(2x+80)\]
\[x^2=2x+80\]
\[x^2-2x-80=0\]
\[(x-10)(x+8)=0\] so
\[x=10\] only because you cannot take the log of a negative number. so discard -8

Answer 4

Actually -8 is extraneous because you can't have a negative log

Answer 5

what edge said