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

solve for "c"