Solution assignment 10 Logarithmic functions and graphs

Solve:

$\log_x(3)+\log_x(27)=2$

Of course we have: $x>0$ and $x\neq1$ .

We apply the first logarithm formula:

$\log_x(3)+\log_x(27)=\log_x(81)=2$

$x^2=81$

$x=9$ of $x=-9$

Because $x=-9$ does not meet the requirements above the only solution is:

$x=9$