$\displaystyle log5 (2x+3) = log5 (3)$ log to base 5
If the bases are the same
the definition is
$\displaystyle log_a{a}^b = b$
so $\displaystyle log_5{5}^1 = 1$
$\displaystyle log_{5}5 (2x+3) = log_5{5}(3)$
then $\displaystyle (2x+3) = 3$
$\displaystyle x = 0$
Hello math321I assume that these are logs to base $\displaystyle 10$; in which case:
$\displaystyle \log x +\log(x-21) = 2$Grandad
$\displaystyle \Rightarrow \log x(x-21)= \log 10^2$
$\displaystyle \Rightarrow x^2-21x=100$
$\displaystyle \Rightarrow (x-25)(x+4)=0$
$\displaystyle \Rightarrow x =25, -4$
...
$\displaystyle \log x +\log(x-21) = 2$
$\displaystyle \Rightarrow \log x(x-21)= \log 10^2$
$\displaystyle \Rightarrow x^2-21x=100$
$\displaystyle \Rightarrow (x-25)(x+4)=0$
$\displaystyle \Rightarrow x =25, \textcolor{red}{-4}$
... be sure to check solutions in the original equation to avoid domain issues.