Question

Solve the equation: log(base4)x+log(base4)(x-3)=1

Answer 1

\[log_4(x)+log_4(x-3)=1\]
rewrite as a single logarithm using property of the logs that
\[log(ab)=log(a)+log(b)\] but use it from right to left
\[log_4(x(x-3))=1\]
rewrite in equivalent exponential form using
\[log_b(x)=y\] means \[b^y=x\]
\[x(x-3)=4^1\]
solve the quadratic equation
\[x^2-3x-4=0\]
\[(x-4)(x-1)=0\]
\[x=4\] or \[x=-1\] but you cannot take log of a negative number so only 4 is the answer

Answer 2

thanks!