Question

Log (C+3) base 3 - log (4C-1) base 3 = log 5 base 3

Answer 1

$\bf log_3(c+3)-log_3(4c-1)=log_3(5)\\\\ -----------------------------\\\\ log_{{ a}}\left( \frac{x}{y}\right)\implies log_{{ a}}(x)-log_{{ a}}(y)\qquad thus\\\\ -----------------------------\\\\ log_3\left( \frac{c+3}{4c-1} \right)=log_3(5)\implies \cfrac{c+3}{4c-1}=5$

solve for "c"